Advanced Algebra

Review of Real Numbers and Absolute Value

🧭 Overview

🧠 One-sentence thesis

This section reviews fundamental operations with real numbers—addition, subtraction, multiplication, and division—emphasizing their properties, the importance of order in non-commutative operations, and techniques for working with fractions.

📌 Key points (3–5)

Addition vs subtraction commutativity: addition is commutative (order doesn't matter), but subtraction is not (order matters).
Multiplication vs division commutativity: multiplication is commutative, but division is not; sequential operations must be performed left to right.
Working with fractions: requires finding a common denominator for addition/subtraction; multiplication is straightforward (multiply numerators and denominators).
Common confusion: sequential operations—must replace all operations with addition or subtraction first, then work left to right; for multiplication/division, work left to right as they appear.
Key properties: each operation has identity, inverse, associative, and commutative properties (where applicable).

➕ Addition and Subtraction Properties

🔑 Core properties of addition

The excerpt defines four fundamental properties for any real numbers a, b, and c:

Property	Formula	Meaning
Additive Identity	a + 0 = 0 + a = a	Adding zero leaves the number unchanged
Additive Inverse	a + (−a) = (−a) + a = 0	A number plus its negative equals zero
Associative	(a + b) + c = a + (b + c)	Grouping doesn't affect the sum
Commutative	a + b = b + a	Order doesn't affect the sum

⚠️ Why order matters for subtraction

Addition is commutative: 5 + 10 = 10 + 5 = 15 (order doesn't matter).
Subtraction is NOT commutative: 5 − 10 = −5, but 10 − 5 = 5 (order matters).
Don't confuse: you cannot rearrange subtraction the way you can rearrange addition.

🔄 Sequential operations strategy

General rule: first replace all sequential operations with either addition or subtraction, then perform each operation in order from left to right.

Example from the excerpt: −10 − (−10) + (−5)

Replace −(−) with addition (+): −10 + 10 − 5
Replace +(−) with subtraction (−): already done
Work left to right: 0 − 5 = −5

➗ Adding and subtracting fractions

Common denominator: a denominator that is shared by more than one fraction.

For fractions with common denominator c: a/c + b/c = (a + b)/c and a/c − b/c = (a − b)/c

Finding the least common denominator (LCD):

The LCD is the least common multiple (LCM) of all denominators.
Method 1: List multiples of each denominator until you find the smallest common one.
Method 2 (more efficient): Use prime factorization—the LCM is the product of each prime factor raised to the highest power appearing in any factorization.

Example: For 2/9 − 1/15 + 8/45, the LCD is 45. Convert each fraction to have denominator 45, then perform operations on numerators: (10 − 3 + 8)/45 = 15/45 = 1/3.

📝 Translating English to math

Common phrases:

"The sum of a number and 2" → n + 2
"The difference of 2 and a number" → 2 − n
"2 subtracted from a number" → n − 2 (note the order!)

Example: "What is 8 subtracted from the sum of 3 and 1/2?" → (3 + 1/2) − 8 = 7/2 − 8 = −9/2

✖️ Multiplication and Division Properties

🔑 Core properties of multiplication

For any real numbers a, b, and c:

Property	Formula	Meaning
Zero Factor	a · 0 = 0 · a = 0	Multiplying by zero gives zero
Multiplicative Identity	a · 1 = 1 · a = a	Multiplying by one leaves the number unchanged
Associative	(a · b) · c = a · (b · c)	Grouping doesn't affect the product
Commutative	a · b = b · a	Order doesn't affect the product

⚠️ Why order matters for division

Multiplication is commutative: 5 · 10 = 10 · 5 = 50 (order doesn't matter).
Division is NOT commutative: 5 ÷ 10 = 0.5, but 10 ÷ 5 = 2 (order matters).
Don't confuse: you cannot rearrange division the way you can rearrange multiplication.

🔄 Sequential multiplication and division

Important: when sequential operations involve multiplication and division, order does matter; work the operations from left to right to obtain a correct result.

Example from the excerpt: 10 ÷ (−2)(−5)

Perform division first (left to right): 10 ÷ (−2) = −5
Then multiply: (−5)(−5) = 25
If you multiply first instead, you get the wrong answer.

Sign rules for multiplication:

Positive × Negative = Negative
Negative × Negative = Positive

Example: 5(−3)(−2)(−4) = multiply two at a time from left to right = −120

➗ Multiplying fractions

The product of two fractions is the fraction formed by the product of the numerators and the product of the denominators.

Formula: a/b · c/d = ac/bd

Example: −4/5 · 25/12

Multiply numerators: (−4)(25) = −100
Multiply denominators: (5)(12) = 60
Result: −100/60
Reduce by dividing out common factors: −5/3

Operations with Real Numbers

🧭 Overview

🧠 One-sentence thesis

Understanding the properties of real numbers and performing operations in the correct order—from left to right for sequential operations, and with proper handling of grouping symbols—is essential for simplifying algebraic expressions.

📌 Key points (3–5)

Addition and subtraction are not interchangeable: addition is commutative (order doesn't matter), but subtraction is not; always work left to right after replacing sequential operations.
Multiplication and division order matters: multiplication is commutative, but division is not; perform multiplication and division as they appear from left to right.
Fractions require common denominators: for addition/subtraction, find the least common denominator (LCD) using the least common multiple (LCM) of denominators.
Common confusion—division vs multiplication of fractions: to divide fractions, multiply by the reciprocal of the divisor, not by the divisor itself.
Grouping symbols dictate operation order: perform operations inside the innermost grouping symbols (parentheses, brackets, braces, fraction bars, absolute value) first.

➕ Addition and Subtraction Properties

🔢 Core properties of addition

The excerpt defines four key properties for any real numbers a, b, and c:

Property	Definition	What it means
Additive Identity	a + 0 = 0 + a = a	Adding zero doesn't change the number
Additive Inverse	a + (−a) = (−a) + a = 0	A number plus its negative equals zero
Associative	(a + b) + c = a + (b + c)	Grouping doesn't affect the sum
Commutative	a + b = b + a	Order doesn't affect the sum

⚠️ Why subtraction is different

Addition is commutative: 5 + 10 = 10 + 5 = 15 (order doesn't matter).
Subtraction is NOT commutative: 5 − 10 = −5, but 10 − 5 = 5 (order matters).
General rule: Replace all sequential operations with either addition or subtraction, then perform each operation in order from left to right.
Example: −10 − (−10) + (−5) becomes −10 + 10 − 5, which equals 0 − 5 = −5.

🧮 Working with fractions

Common denominator: a denominator that is shared by more than one fraction.

Least common denominator (LCD): the least common multiple of all the denominators.

For fractions with common denominator c: a/c + b/c = (a + b)/c and a/c − b/c = (a − b)/c.
To find the LCD, list multiples of each denominator or use prime factorization (the LCM is the product of each prime factor raised to the highest power).
Example: For 2/9 − 1/15 + 8/45, the LCD is 45; convert each fraction to denominator 45, then compute (10 − 3 + 8)/45 = 15/45 = 1/3.

📝 Translating word problems

Common phrases and their meanings:

"The sum of a number and 2" → n + 2
"The difference of 2 and a number" → 2 − n
"8 subtracted from a number" → n − 2 (not 2 − n; order matters!)
Example: "What is 8 subtracted from the sum of 3 and 1/2?" → (3 + 1/2) − 8 = 7/2 − 8 = −9/2.

✖️ Multiplication and Division Properties

🔢 Core properties of multiplication

The excerpt defines four key properties for any real numbers a, b, and c:

Property	Definition	What it means
Zero Factor	a · 0 = 0 · a = 0	Multiplying by zero gives zero
Multiplicative Identity	a · 1 = 1 · a = a	Multiplying by one doesn't change the number
Associative	(a · b) · c = a · (b · c)	Grouping doesn't affect the product
Commutative	a · b = b · a	Order doesn't affect the product

⚠️ Why division is different

Multiplication is commutative: 5 · 10 = 10 · 5 = 50 (order doesn't matter).
Division is NOT commutative: 5 ÷ 10 = 0.5, but 10 ÷ 5 = 2 (order matters).
Critical rule: When sequential operations involve multiplication and division, work the operations from left to right to obtain a correct result.
Example: 10 ÷ (−2) · (−5) must be done left to right: first 10 ÷ (−2) = −5, then (−5) · (−5) = 25. Doing it in the wrong order gives an incorrect result.

🔄 Sign rules for products

Positive × Negative = Negative
Negative × Negative = Positive
Example: 5(−3)(−2)(−4) = (−15)(−2)(−4) = (30)(−4) = −120.

🍰 Multiplying fractions

To multiply fractions, multiply the numerators and multiply the denominators: a/b · c/d = ac/bd.

Reduce by dividing out common factors.
Example: −4/5 · 25/12 = −(4 · 25)/(5 · 12) = −(4 · 25)/(5 · 12) = −5/3 (after canceling common factors).

🔁 Reciprocals and dividing fractions

Reciprocals: two real numbers whose product is 1.

a/b and b/a are reciprocals because a/b · b/a = ab/ab = 1.
Examples of reciprocals: 2/3 and 3/2; 7 and 1/7; −4/5 and −5/4.
To divide fractions: multiply the dividend by the reciprocal of the divisor.
Formula: a/b ÷ c/d = a/b · d/c = ad/bc.
Example: 5/4 ÷ 3/5 · 1/2 = 5/4 · 5/3 · 1/2 = (5 · 5 · 1)/(4 · 3 · 2) = 25/24 (work left to right).
Don't confuse: Division requires the reciprocal; you cannot simply multiply by the divisor as-is.

🗂️ Grouping Symbols and Order of Operations

📦 What are grouping symbols

Grouping symbols: parentheses ( ), brackets [ ], braces { }, and the fraction bar are the common symbols used to group expressions and mathematical operations within a computation.

Absolute value bars also function as grouping symbols.
All grouping symbols have the same order of precedence.

🎯 How to handle grouping symbols

Perform operations inside the innermost grouping symbol or absolute value first.
Work outward from the innermost grouping to the outermost.
The excerpt emphasizes that grouping symbols "help tell us which operations to perform first" in computations involving more than one operation.

Square and Cube Roots of Real Numbers

🧭 Overview

🧠 One-sentence thesis

Exponential notation compactly represents repeated multiplication, and understanding the base and exponent—especially with negative bases—is essential for correctly applying the order of operations in algebraic calculations.

📌 Key points (3–5)

Reciprocals: Two numbers whose product is 1; dividing fractions requires multiplying by the reciprocal of the divisor.
Exponential notation: A compact way to write repeated multiplication, where the base is the factor and the exponent tells how many times to repeat it.
Common confusion: Negative bases with parentheses vs. without—parentheses determine whether the negative sign is part of the base or applied after exponentiation.
Order of operations: Grouping symbols (innermost first), then exponents, then multiplication/division (left to right), finally addition/subtraction (left to right).
Even vs. odd exponents: A negative base raised to an even exponent yields a positive result; an odd exponent yields a negative result.

🔄 Reciprocals and division

🔄 What reciprocals are

Reciprocals: Two real numbers whose product is 1.

If two fractions multiply to equal 1, they are reciprocals.
Example: 2/3 and 3/2 are reciprocals because (2/3) · (3/2) = 1.
The general form: a/b and b/a are reciprocals because (a/b) · (b/a) = ab/ab = 1.

➗ Why reciprocals matter for division

Dividing fractions requires multiplying the dividend by the reciprocal of the divisor.
The general rule: (a/b) ÷ (c/d) = (a/b) · (d/c) = ad/bc.
Example: To divide 5/4 by 3/5, multiply 5/4 by the reciprocal 5/3, then continue operations left to right.
Don't confuse: Division is not commutative—order matters, unlike multiplication.

📐 Exponential notation basics

📐 What exponents represent

Exponential notation: The compact notation a^n used when a factor a is repeated n times.

The base is the factor being repeated.
The exponent (a positive integer) indicates how many times the base is used as a factor.
Example: 5 · 5 · 5 · 5 = 5^4, where 5 is the base and 4 is the exponent.

🔢 Special names for exponents

Exponent	Name	Example
2	Square	5^2 = 5 · 5 = 25 ("5 squared")
3	Cube	5^3 = 5 · 5 · 5 = 125 ("5 cubed")
n > 3	nth power	Read as "a raised to the nth power"

The base can be any real number, including fractions and negative numbers.

⚠️ Negative bases and parentheses

⚠️ The critical difference

The excerpt emphasizes that parentheses determine whether the negative sign is part of the base:

Expression	Base	Calculation	Result
(−3)^4	−3	(−3)(−3)(−3)(−3)	+81
−3^4	3	−1 · 3 · 3 · 3 · 3	−81
(−3)^3	−3	(−3)(−3)(−3)	−27
−3^3	3	−1 · 3 · 3 · 3	−27

With parentheses: The negative number is the base.
Without parentheses: Only the positive number is the base; the negative sign is applied after exponentiation.

🔀 Even vs. odd exponents

Negative base with even exponent: Result is positive.
- Example: (−2)^4 = (−2)(−2)(−2)(−2) = 16
Negative base with odd exponent: Result is negative.
- Example: (−2/3)^3 = (−2/3)(−2/3)(−2/3) = −8/27
Don't confuse: Without parentheses, −2^4 = −16 because the base is 2, not −2.

🎯 Order of operations

🎯 The sequence to follow

Innermost grouping symbols first: Parentheses, brackets, braces, fraction bars, and absolute values all have the same precedence.
Evaluate all exponents.
Multiplication and division from left to right.
Addition and subtraction from left to right.

📦 Grouping symbols

Grouping symbols: Parentheses, brackets, braces, and the fraction bar are common symbols used to group expressions and mathematical operations within a computation.

All grouping symbols have the same order of precedence.
Perform operations inside the innermost grouping first.
Absolute value bars also act as grouping symbols.

↔️ Left-to-right operations

Multiplication and division should be worked from left to right as they appear.
It is often reasonable to perform division before multiplication, depending on the order in the expression.
Example: In 5^3 − 24 ÷ 6 · (1/2) + 2, after evaluating 5^3 = 125, perform 24 ÷ 6 = 4, then 4 · (1/2) = 2, working left to right.
Don't confuse: Multiplying first (ignoring left-to-right order) leads to incorrect results.

🧮 Working with absolute values

Assign absolute value the same precedence as parentheses.
Perform operations inside the absolute value first before applying the absolute value itself.
Example: In 7 − 5|−2^2 + (−3)^2|, first calculate inside: −4 + 9 = 5, then |5| = 5, then 7 − 5 · 5 = 7 − 25 = −18.

🔑 Key takeaways from the excerpt

🔑 Operation properties

Addition is commutative; subtraction is not.
Multiplication is commutative; division is not.
Adding or subtracting fractions requires a common denominator; multiplying or dividing fractions does not.

🔑 Ensuring correct results

To ensure a single correct result, always follow the order of operations.
Work one operation at a time to reduce mistakes.
Correctly identify the base when squaring or raising to any power—parentheses are crucial for negative bases.

Operations with Real Numbers and Roots

Algebraic Expressions and Formulas

🧭 Overview

🧠 One-sentence thesis

Following the correct order of operations—parentheses, exponents, multiplication/division, then addition/subtraction—is essential for evaluating expressions with real numbers, and understanding square roots requires distinguishing between the principal (positive) root and the negative root.

📌 Key points (3–5)

Order of operations: perform operations in parentheses first, then exponents, then multiplication/division left to right, finally addition/subtraction left to right.
Square root definition: a square root of a number is a number that when multiplied by itself yields the original number; every positive real number has two square roots (one positive, one negative).
Radical sign notation: the radical sign √ denotes the principal (non-negative) square root; a negative sign in front (−√) denotes the negative square root.
Common confusion: the square root of a squared negative number is positive, not negative—√(−3)² = √9 = 3, not −3, because the radical always denotes the principal root.
Perfect squares vs non-perfect squares: if the radicand can be factored as a square of another number, the square root is apparent; otherwise, the square root of a positive non-perfect-square integer is irrational.

🔢 Order of operations

🔢 The sequence to follow

The excerpt states the order clearly:

First, perform operations in the innermost parentheses or groupings.
Next, simplify all exponents.
Perform multiplication and division operations from left to right.
Finally, perform addition and subtraction operations from left to right.

This sequence ensures consistent evaluation of expressions.

🧮 Why it matters

Without a standard order, the same expression could yield different results.
Example: 5 − 3 × 4 could be (5 − 3) × 4 = 8 or 5 − (3 × 4) = −7; the correct order gives 5 − 12 = −7.
The excerpt includes many practice problems (e.g., "5 − 3 [3 (2 − 3²) + (−3)²]") that require applying this order step by step.

🧩 Square roots: definition and notation

🧩 What a square root is

A square root of a number is a number that when multiplied by itself yields the original number.

Example: 4 is a square root of 16 because 4² = 16.
Also, −4 is a square root of 16 because (−4)² = 16.
Every positive real number has two square roots: one positive, one negative.
Zero is the only real number with exactly one square root: √0 = 0.

📐 The radical sign and principal root

The excerpt introduces two notations:

Notation	Meaning	Example
√	Principal (non-negative) square root	√16 = 4
−√	Negative square root	−√16 = −4

The radicand is the number inside the radical sign.
The radical sign √ always denotes the non-negative root.

⚠️ Common confusion: √(−3)² ≠ −3

The excerpt emphasizes:

√(−3)² = √9 = 3, not −3.
Why? Because the radical denotes the principal (positive) square root.
The property √(a²) = a requires a ≥ 0.
Don't confuse: squaring a negative number gives a positive result, and the square root of that positive result is positive.

🔍 Perfect squares and exact roots

🔍 When the square root is apparent

If the radicand can be factored as the square of another nonzero number, the square root is straightforward.

The excerpt gives the property:

√(a²) = a, if a ≥ 0.

📋 Examples from the excerpt

Expression	Factorization	Result
√121	√(11²)	11
√0.25	√(0.5²)	0.5
√(4/9)	√((2/3)²)	2/3
−√64	−√(8²)	−8
−√1	−√(1²)	−1

These are all perfect squares, so the roots are rational and exact.

🌀 Non-perfect squares

If a positive integer is not a perfect square, its square root will be irrational.
Example: the excerpt mentions √5, which cannot be expressed as a simple fraction.
The text cuts off here, but the implication is that such roots require approximation or simplification techniques.

🧮 Cube roots (brief mention)

🧮 Definition

The excerpt's learning objectives mention cube roots but do not provide detailed content in the visible portion.

A cube root of a number is a number that when multiplied by itself three times yields the original number.
Example: the cube root of 8 is 2, because 2³ = 8.
The excerpt promises to cover exact and approximate values and simplification, but those sections are not included in the provided text.

📏 Additional concepts

📏 Distance formula on a number line

The excerpt introduces:

d = |b − a| gives the distance between any two points on a number line.
Example: distance between 10 and 15 is |15 − 10| = 5 units.
Example: distance between −5 and −25 is |−25 − (−5)| = |−20| = 20 units.

🔄 Reciprocals

The reciprocal of a number is 1 divided by that number.
Example: reciprocal of 1/3 is 3; reciprocal of −3/4 is −4/3.
The reciprocal of a (where a ≠ 0) is 1/a.
Zero has no reciprocal (division by zero is undefined).

📊 Real-world applications

The excerpt includes word problems:

Calculating pay with overtime (time and a half).
Splitting a distance or length into equal parts.
Finding averages.
Temperature range (hottest minus coldest).

These illustrate how operations with real numbers apply to practical scenarios.

Square and Cube Roots of Real Numbers

Rules of Exponents and Scientific Notation

🧭 Overview

🧠 One-sentence thesis

Square roots and cube roots extract factors from numbers, and simplifying them requires identifying perfect-power factors and applying product/quotient rules to express radicals in their simplest form.

📌 Key points (3–5)

Principal vs negative square root: the radical sign √ denotes only the non-negative (principal) square root; use −√ for the negative root.
Square roots of negatives are undefined (for now): any real number squared is positive, so square roots of negative numbers are not real numbers in this context.
Cube roots differ from square roots: cube roots exist for all real numbers (positive, negative, and zero) because multiplying three equal factors preserves sign.
Common confusion—perfect vs imperfect powers: if the radicand is a perfect square/cube, the root is rational; otherwise, it is irrational and must be simplified by factoring out the largest perfect power.
Simplification uses product/quotient rules: split the radicand into a perfect-power factor and a remaining factor, then extract the root of the perfect power.

🔢 Square roots: definition and properties

🔢 What a square root is

A square root of a number is a number that when multiplied by itself yields the original number.

Every positive real number has two square roots: one positive, one negative.
Example: both 4 and −4 are square roots of 16, because 4² = 16 and (−4)² = 16.
Zero has exactly one square root: 0.

✅ Principal (non-negative) square root

The radical sign √ denotes the principal square root, which is always non-negative.
To denote the negative square root, write −√.
Example: √16 = 4 (principal), −√16 = −4 (negative).
Don't confuse: √(−3)² ≠ −3; instead, √(−3)² = √9 = 3, because the radical always returns the non-negative result.

❌ Square roots of negative numbers

The square of any real number is positive, so no real number squared equals a negative number.
Example: to find √(−9), you would need a number whose square is −9, but (3)² = 9 and (−3)² = 9.
Conclusion: √(−9) is not a real number (it is undefined in this course section).

🔍 Perfect squares vs irrational roots

If the radicand is a perfect square (e.g., 16 = 4²), the square root is rational.
If not, the square root is irrational.
Example: √5 is irrational; it lies between √4 = 2 and √9 = 3, approximately 2.236.

🧊 Cube roots: definition and properties

🧊 What a cube root is

A cube root of a number is a number that when multiplied by itself three times yields the original number.

Denoted by ∛ (the index 3 indicates cube root).
Example: ∛8 = 2, because 2³ = 8.

➕➖ Cube roots of positive and negative numbers

Any real number has exactly one real cube root (no "principal root" technicality).
Positive radicand → positive cube root.
Negative radicand → negative cube root.
Example: ∛(−8) = −2, because (−2)³ = −8.
Zero: ∛0 = 0.

🔍 Perfect cubes vs irrational cube roots

If the radicand is a perfect cube (e.g., 27 = 3³), the cube root is rational.
Otherwise, the cube root is irrational.
Example: ∛2 ≈ 1.260 (irrational).

📐 General property

For any real number a:

∛(a³) = a (no restriction on sign).

🛠️ Simplifying radicals: product and quotient rules

🛠️ What "simplified radical" means

A simplified radical is one where the radicand does not consist of any factors that can be written as perfect powers of the index.

For square roots: factor out the largest perfect square.
For cube roots: factor out the largest perfect cube.

📏 Product rule for radicals

Product rule: ⁿ√(A · B) = ⁿ√A · ⁿ√B

Use this to split a radicand into a perfect-power factor and a remaining factor.
Example (square root): √12 = √(4 · 3) = √4 · √3 = 2√3.
Example (cube root): ∛80 = ∛(8 · 10) = ∛8 · ∛10 = 2∛10.

➗ Quotient rule for radicals

Quotient rule: ⁿ√(A/B) = ⁿ√A / ⁿ√B (where B ≠ 0)

Use this to simplify fractions under a radical.
Example: √(108/169) = √108 / √169 = (√(36·3)) / 13 = (6√3) / 13.

🔧 Step-by-step simplification process

Factor the radicand into prime factors or identify the largest perfect power.
Apply the product/quotient rule to separate the perfect-power factor.
Extract the root of the perfect-power factor.
Leave the remaining factor under the radical.

Example (square root): Simplify √135.

Factor: 135 = 9 · 15 = 3² · 15.
Apply product rule: √135 = √(9 · 15) = √9 · √15 = 3√15.

Example (cube root): Simplify ∛162.

Factor: 162 = 27 · 6 = 3³ · 6.
Apply product rule: ∛162 = ∛(27 · 6) = ∛27 · ∛6 = 3∛6.

🧮 Worked examples and common patterns

🧮 Simplifying square roots with coefficients

Example: Simplify −5√162.

Factor 162: 162 = 81 · 2 = 9² · 2.
Apply product rule: √162 = √(81 · 2) = √81 · √2 = 9√2.
Multiply by coefficient: −5 · 9√2 = −45√2.

🧮 Simplifying cube roots of fractions

Example: Simplify −∛(16/343).

Factor numerator: 16 = 8 · 2 = 2³ · 2.
Factor denominator: 343 = 7³.
Apply quotient rule: ∛(16/343) = ∛16 / ∛343 = (∛(8·2)) / 7 = (2∛2) / 7.
Apply negative sign: −(2∛2) / 7.

🧮 Simplifying powers of square roots

Property: (√a)² = a (if a ≥ 0).

Example: (√10)² = 10.
This also works in reverse: √(a²) = a (if a ≥ 0).
Don't confuse: the order of squaring and taking the root does not matter for non-negative numbers.

📐 Pythagorean theorem (introduction)

📐 Right triangle terminology

A right triangle is a triangle where one of the angles measures 90°.

Hypotenuse: the longest side, opposite the right angle.
Legs: the other two sides.

📐 The Pythagorean theorem

The Pythagorean theorem states that given any right triangle with legs measuring a and b units, the square of the measure of the hypotenuse c is equal to the sum of the squares of the legs.

Formula: c² = a² + b².
This theorem connects geometry to square roots: to find the hypotenuse, take the square root of (a² + b²).
(The excerpt ends here; further applications are not provided.)

📋 Comparison table: square roots vs cube roots

Feature	Square roots	Cube roots
Notation	√ (index 2, usually omitted)	∛ (index 3 must be shown)
Number of real roots	Two (positive and negative) for positive radicands; zero for negative radicands	Exactly one for any real radicand
Negative radicands	Not real numbers (undefined in this context)	Real numbers (negative cube root)
Principal root	√ denotes only the non-negative root	No "principal" distinction needed
Simplification	Factor out largest perfect square	Factor out largest perfect cube
Example (positive)	√16 = 4, −√16 = −4	∛8 = 2
Example (negative)	√(−9) is not real	∛(−8) = −2

Square and Cube Roots of Real Numbers

Polynomials and Their Operations

🧭 Overview

🧠 One-sentence thesis

Simplifying square and cube roots requires identifying perfect square or cube factors of the radicand, and the Pythagorean theorem provides a necessary and sufficient condition for determining whether a triangle is a right triangle.

📌 Key points (3–5)

Square roots: the principal square root of a positive number is the positive number that when squared gives the original; square roots of negative numbers are undefined.
Cube roots: every real number has exactly one real cube root (a number that when cubed gives the original), and negative numbers can have cube roots.
Simplification strategy: find the largest perfect square (or perfect cube) factor of the radicand, apply the product or quotient rule, then simplify.
Common confusion: square vs cube roots—square roots of negatives are undefined, but cube roots of negatives exist and are negative.
Pythagorean theorem: states that a squared plus b squared equals c squared if and only if the triangle is a right triangle (necessary and sufficient condition).

🔢 Square roots

🔢 Definition and principal root

The square root of a number is a number that when squared results in the original number. The principal square root of a positive real number is the positive square root.

For any positive number, there are technically two square roots (one positive, one negative), but the principal root is always the positive one.
The square root of a negative number is currently left undefined (not a real number).
Example: the square root of 81 is 9, because 9 times 9 equals 81.

⚠️ Negative numbers and square roots

You cannot take the square root of a negative number within the real number system.
Example: the square root of negative 16 is not a real number.
Don't confuse: the square root of (negative 5) squared equals 5, because you square first (giving positive 25), then take the square root.

🧊 Cube roots

🧊 Definition and uniqueness

The cube root of a number is a number that when cubed results in the original number. Every real number has only one real cube root.

Unlike square roots, cube roots exist for all real numbers, including negatives.
The cube root of a negative number is negative.
Example: the cube root of 64 is 4 (because 4 times 4 times 4 equals 64); the cube root of negative 27 is negative 3.

🔄 Key difference from square roots

Feature	Square roots	Cube roots
Negative radicands	Undefined (not real)	Defined (negative result)
Number of real roots	Two (but principal is positive)	Exactly one
Example	Square root of negative 1 is undefined	Cube root of negative 1 is negative 1

🛠️ Simplification techniques

🛠️ Product and quotient rules

To simplify, identify the largest perfect square factor (for square roots) or perfect cube factor (for cube roots) of the radicand.
Apply the product rule: the root of a product equals the product of the roots.
Apply the quotient rule: the root of a quotient equals the quotient of the roots.

📐 Simplifying square roots

Look for perfect square factors (1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, etc.).
Example: to simplify the square root of 80, notice that 80 equals 16 times 5. The square root of 80 equals the square root of 16 times the square root of 5, which simplifies to 4 times the square root of 5.
Example: to simplify the square root of 162, factor as 81 times 2, giving 9 times the square root of 2.

🧊 Simplifying cube roots

Look for perfect cube factors (1, 8, 27, 64, 125, 216, 343, 512, etc.).
Example: to simplify the cube root of 80, notice that 80 equals 8 times 10. The cube root of 80 equals the cube root of 8 times the cube root of 10, which simplifies to 2 times the cube root of 10.
Example: to simplify the cube root of 162, factor as 27 times 6 (since 162 equals 3 to the fourth times 2, which is 27 times 6), giving 3 times the cube root of 6.

🔁 Exponent and root relationship

For any nonnegative number a: the square root of a squared equals a, and the square of the square root of a also equals a.
Example: the square root of 81 equals 9, and 9 squared equals 81; also, the square of the square root of 10 equals 10.
This property holds regardless of whether you apply the exponent first or the root first.

📐 Pythagorean theorem

📐 The theorem

The Pythagorean theorem states that given any right triangle with legs measuring a and b units, the square of the measure of the hypotenuse c is equal to the sum of the squares of the measures of the legs: a squared plus b squared equals c squared.

A right triangle is a triangle where one angle measures 90 degrees.
The hypotenuse is the longest side, opposite the right angle.
The other two sides are called legs.
In other words, the hypotenuse equals the square root of the sum of the squares of the legs.

✅ Necessary and sufficient condition

The theorem provides both a necessary and sufficient condition: a squared plus b squared equals c squared if and only if the triangle is a right triangle.
This means: if you can show that the sum of the squares of the legs equals the square of the hypotenuse, then the triangle must be a right triangle.
Example: a triangle with legs 1 cm and 2 cm and hypotenuse the square root of 5 cm is a right triangle, because 1 squared plus 2 squared equals 1 plus 4 equals 5, which equals the square of the square root of 5.

🧮 Applications

Diagonal of a square: if a square has sides measuring 5 units, the diagonal forms an isosceles right triangle with two legs of 5 units each. Using the theorem: c equals the square root of (5 squared plus 5 squared) equals the square root of 50 equals 5 times the square root of 2 units.
Verifying right triangles: given three side lengths, check whether a squared plus b squared equals c squared to determine if the triangle is a right triangle.
Example: a triangle with a equals 3, b equals 7, and c equals 10 is not a right triangle, because 9 plus 49 equals 58, which does not equal 100.

Algebraic Expressions and Formulas

Solving Linear Equations

🧭 Overview

🧠 One-sentence thesis

Algebraic expressions generalize arithmetic operations using variables, and by applying the distributive property and combining like terms we can simplify these expressions and use them in reusable formulas that model real-world applications.

📌 Key points (3–5)

What algebraic expressions are: combinations of variables, numbers, and mathematical operations that generalize arithmetic.
How to simplify: apply the distributive property to remove parentheses, then combine like terms by adding/subtracting coefficients.
Evaluating expressions: substitute given values for variables (best practice: use parentheses first) and simplify using order of operations.
Common confusion: like terms must have exactly the same variable parts (same variables with same exponents)—only coefficients change when combining.
Why formulas matter: algebraic expressions allow us to create reusable formulas for common applications (geometry, motion, interest, etc.).

🧩 Building blocks of expressions

🧩 What makes up an expression

Algebraic expressions: Combinations of variables and numbers along with mathematical operations used to generalize specific arithmetic operations.

Variables are letters (like x, y) that represent numbers.
The excerpt emphasizes that expressions are generalizations of arithmetic operations.

📦 Terms, factors, and coefficients

Terms: Components of an algebraic expression separated by addition operators.

Factors: Components of a term separated by multiplication operators.

Coefficient: The numerical factor of a term.

Variable part of a term: All the variable factors with their exponents.

Constant term: A term written without a variable factor.

Example from the excerpt: In x²y² + 6xy − 3, there are three terms.
First term x²y² has coefficient 1 and variable part x²y².
Second term 6xy has coefficient 6 and variable part xy.
Third term −3 is a constant (coefficient −3, no variable part).
Don't confuse: the coefficient includes the sign; −b² means coefficient is −1, not 1.

🔧 The distributive property

🔧 What it states

Distributive property: Given any real numbers a, b, and c, then a(b + c) = ab + ac or (b + c)a = ba + ca.

This property is "one that we apply often when simplifying algebraic expressions."
It allows us to multiply every term inside parentheses by the factor outside.

🔧 When to use it

The excerpt shows two approaches to 2(5 − 3):

Working parentheses first: 2(2) = 4
Using distributive property: 2·5 − 2·3 = 10 − 6 = 4

Key principle: "If the contents of the parentheses can be simplified we should do that first. On the other hand, when the contents of parentheses cannot be simplified any further, we multiply every term within it by the factor outside of it using the distributive property."

Example from excerpt: 5(−2a + 5b) − 2c becomes −10a + 25b − 2c by multiplying only the grouped terms.

🔗 Combining like terms

🔗 What are like terms

Like terms (or similar terms): Constant terms or terms whose variable parts have the same variables with the same exponents.

The variable factors and their exponents must be exactly the same.
Only the coefficients can differ.
Constant terms are always like terms with each other.

🔗 How to combine them

The excerpt shows: "If the variable parts of terms are exactly the same, then we can add or subtract the coefficients to obtain the coefficient of a single term with the same variable part."

Examples from the excerpt:

5x + 7x = (5 + 7)x = 12x
4x² + 5x² − 7x² = (4 + 5 − 7)x² = 2x²
12x²y³ + 3x²y³ = 15x²y³

Important: "The variable factors and their exponents do not change."

🔗 Simplifying expressions

Simplifying the expression: The process of combining like terms until the expression contains no more similar terms.

Example from excerpt: x² − 10x + 8 + 5x² − 6x − 1 simplifies to 6x² − 16x + 7 by identifying and combining like terms.

🔢 Evaluating expressions

🔢 What evaluating means

Evaluating: The process of performing the operations of an algebraic expression for given values of the variables.

Substitute: The act of replacing a variable with an equivalent quantity.

🔢 Best practice approach

The excerpt emphasizes: "To avoid common errors, it is a best practice to first replace all variables with parentheses, and then replace, or substitute, the appropriate given value."

Three-step process shown:

Replace variables with parentheses
Substitute in the appropriate values
Simplify using order of operations

Example from excerpt: Evaluating 5x − 2 where x = 2/3:

5( ) − 2
5(2/3) − 2
10/3 − 2 = 4/3

🔢 Multiple variables

When expressions involve more than one variable, substitute all given values.

Example from excerpt: a³ − 8b³ where a = −1 and b = 1/2 evaluates to −2 after careful substitution and simplification.

Don't confuse: Use parentheses to avoid sign errors, especially with negative values and exponents.

📐 Using formulas

📐 What formulas are

Formulas: A reusable mathematical model using algebraic expressions to describe a common application.

The excerpt states: "The main difference between algebra and arithmetic is the organized use of variables. This idea leads to reusable formulas."

📐 Geometric formulas

The excerpt provides formulas for:

Shape	Formula type	Variables
Right circular cone	Volume: V = (1/3)πr²h	r = radius, h = height
Sphere	Volume: V = (4/3)πr³	r = radius
Rectangle	Perimeter (P), Area (A)	dimensions

Example from excerpt: A cone with radius 3 meters and height 5 meters has volume V = 15π cubic meters (approximately 47.1 cubic meters using π ≈ 3.14).

📐 Application formulas

Uniform motion:

Uniform motion: The distance D after traveling at an average rate r for some time t can be calculated using the formula D = rt.

Read as "distance equals rate times time."
Example from excerpt: 2.5 hours at 66 mph gives D = 66 · 2.5 = 165 miles.

Simple interest:

Simple interest: Modeled by the formula I = prt, where p represents the principal amount invested at an annual interest rate r for t years.

Example from excerpt: $1,250 invested for 2 years at 3.75% annual rate earns I = (1,250)(0.0375)(2) = $93.75.
Don't confuse: Convert percentages to decimals before substituting (3¾% = 0.0375).

Algebraic Expressions and Formulas

Solving Linear Inequalities with One Variable

🧭 Overview

🧠 One-sentence thesis

Algebraic expressions combine variables, numbers, and operations to generalize arithmetic, and understanding their structure—terms, coefficients, and like terms—enables simplification through the distributive property and evaluation by substitution.

📌 Key points (3–5)

What an algebraic expression is: combinations of variables (letters representing numbers) and numbers joined by mathematical operations.
Structure of expressions: terms are separated by addition; factors are separated by multiplication; the numerical factor of a term is its coefficient.
Distributive property: multiply every term inside parentheses by the factor outside to remove parentheses when simplification inside is not possible.
Like terms vs unlike terms: like terms have exactly the same variable part (same variables with same exponents); only like terms can be combined by adding/subtracting coefficients.
Evaluating expressions: replace variables with given values (using parentheses first) and simplify using order of operations.

🧩 Building blocks of expressions

🧩 What an algebraic expression is

Algebraic expressions: combinations of variables and numbers along with mathematical operations used to generalize specific arithmetic operations.

Variables are letters (commonly x and y, but also Greek letters or subscripted letters like x₁, x₂) that represent unknown or changing numbers.
Example expressions with one variable x: 2x + 3, x² − 9, 1/x + x/(x+2), 3√x + √x.
Some letters are reserved for constants (π ≈ 3.14159, e ≈ 2.71828) and do not change.

🔢 Terms, factors, and coefficients

Terms: components of an algebraic expression separated by addition operators.

Factors: components of a term separated by multiplication operators.

Coefficient: the numerical factor of a term.

Terms are separated by addition (or subtraction, which is addition of a negative).
Within each term, factors are separated by multiplication.
Example: x²y² + 6xy − 3 has three terms.
- First term x²y² has coefficient 1 (since x²y² = 1·x·x·y·y).
- Second term 6xy has coefficient 6.
- Third term −3 is a constant term (no variable factor).

Term	Coefficient	Variable Part
x²y²	1	x²y²
6xy	6	xy
−3	−3	(none)

📦 Variable part and constant terms

Variable part of a term: all the variable factors with their exponents.

Constant term: a term written without a variable factor.

The variable part does not include the coefficient.
Constant terms do not change; variables represent quantities that may change.
Example: in 10a² − 5ab − b², the term −b² should be thought of as −1·b², so coefficient is −1 and variable part is b².

🔄 The distributive property

🔄 What the distributive property states

Distributive property: given any real numbers a, b, and c, a(b + c) = ab + ac or (b + c)a = ba + ca.

This property applies because variables represent real numbers.
It allows multiplying a factor outside parentheses by every term inside.
When the contents of parentheses cannot be simplified further, use the distributive property to remove parentheses.

🛠️ How to apply the distributive property

Multiply the factor outside by every term inside the parentheses.
Example: 5(−2a + 5b) − 2c
- Distribute 5: 5·(−2a) + 5·5b − 2c = −10a + 25b − 2c.
- The −2c is not inside the parentheses, so it is not affected.
Example: (3x − 4y + 1)·3
- Distribute 3: 3x·3 − 4y·3 + 1·3 = 9x − 12y + 3.
Don't confuse: only multiply terms inside the parentheses by the outside factor; terms outside are left alone.

🧮 When to use the distributive property

If parentheses can be simplified first (e.g., 2(5 − 3) = 2(2) = 4), do that.
If parentheses cannot be simplified (e.g., contain variables), apply the distributive property to remove them.
Example comparison:
- Working parenthesis first: 2(5 − 3) = 2(2) = 4.
- Using distributive property: 2(5 − 3) = 2·5 − 2·3 = 10 − 6 = 4.
- Both give the same result.

🔗 Combining like terms

🔗 What like terms are

Like terms (or similar terms): constant terms or terms whose variable parts have the same variables with the same exponents.

The variable parts must be exactly the same (same variables, same exponents).
Constant terms are always like terms with each other.
Example: 5x and 7x are like terms (both have variable part x).
Example: 4x² and 5x² and −7x² are like terms (all have variable part x²).
Example: 12x²y³ and 3x²y³ are like terms (both have variable part x²y³).

➕ How to combine like terms

Combining like terms: adding or subtracting like terms within an algebraic expression to obtain a single term with the same variable part.

Add or subtract the coefficients only; the variable part stays the same.
Example: 5x + 7x = (5 + 7)x = 12x.
Example: 4x² + 5x² − 7x² = (4 + 5 − 7)x² = 2x².
Example: 12x²y³ + 3x²y³ = 15x²y³.
Don't confuse: the variable factors and their exponents do not change when combining.

🧹 Simplifying expressions

Simplifying the expression: combining like terms in an expression until it contains no more similar terms.

Identify all like terms and add their coefficients.
Example: x² − 10x + 8 + 5x² − 6x − 1
- Like terms: x² and 5x² → (1 + 5)x² = 6x².
- Like terms: −10x and −6x → (−10 − 6)x = −16x.
- Like terms: 8 and −1 → 8 − 1 = 7.
- Simplified: 6x² − 16x + 7.
Example with distributive property first: a²b² − ab − 2(2a²b² − 5ab + 1)
- Distribute −2: a²b² − ab − 4a²b² + 10ab − 2.
- Combine like terms: (1 − 4)a²b² + (−1 + 10)ab − 2 = −3a²b² + 9ab − 2.

🔢 Evaluating algebraic expressions

🔢 What evaluating means

Evaluating: the process of performing the operations of an algebraic expression for given values of the variables.

An algebraic expression generalizes arithmetic operations.
When a problem assigns specific values to variables, replace the variables with those numbers and compute using order of operations.

🔄 How to substitute values safely

Substitute: the act of replacing a variable with an equivalent quantity.

Best practice: first replace all variables with parentheses, then substitute the given values inside the parentheses.
This avoids common errors, especially with negative numbers or fractions.
Example: evaluate 5x − 2 where x = 2/3.
- Replace variable with parentheses: 5( ) − 2.
- Substitute: 5(2/3) − 2 = 10/3 − 2 = 10/3 − 6/3 = 4/3.
Example: evaluate y² − y − 6 where y = −4.
- Replace: ( )² − ( ) − 6.
- Substitute: (−4)² − (−4) − 6 = 16 + 4 − 6 = 14.

🧮 Evaluating with multiple variables

Expressions can involve more than one variable.
Substitute all given values and simplify using correct order of operations.
Example: evaluate a³ − 8b³ where a = −1 and b = 1/2.
- Replace: ( )³ − 8( )³.
- Substitute: (−1)³ − 8(1/2)³.
- Simplify: −1 − 8(1/8) = −1 − 1 = −2.
Don't confuse: follow order of operations (exponents before multiplication/subtraction).

Relations, Graphs, and Functions

🧭 Overview

🧠 One-sentence thesis

The rectangular coordinate system allows us to visualize relations as graphs, and among all relations, functions are special because each input corresponds to exactly one output, which we can verify using the vertical line test and express using function notation.

📌 Key points (3–5)

Rectangular coordinate system: Uses two perpendicular number lines (x-axis and y-axis) to locate points as ordered pairs (x, y) in a plane divided into four quadrants.
Relations vs. functions: A relation is any set of ordered pairs; a function is a special relation where each x-value corresponds to exactly one y-value.
Domain and range: The domain is the set of all x-values (inputs); the range is the set of all y-values (outputs).
Common confusion—vertical line test: If any vertical line crosses a graph more than once, the relation is not a function; one x-value would correspond to multiple y-values.
Function notation f(x): Reads "f of x" and means the output value when x is the input; f(x) and y are interchangeable.

📐 The rectangular coordinate system

📐 Components and structure

Rectangular coordinate system: A system with two number lines at right angles specifying points in a plane using ordered pairs (x, y).

The x-axis is the horizontal number line.
The y-axis is the vertical number line.
The origin is where the axes cross, at point (0, 0).
The plane is the flat surface defined by these axes.
Quadrants are the four regions (I, II, III, IV) created by the axes.

📍 Ordered pairs

Ordered pair: Pairs (x, y) that identify position relative to the origin on a rectangular coordinate plane.

The first number is the x-coordinate (horizontal position).
The second number is the y-coordinate (vertical position).
Example: (−4, 3) means 4 units left of the origin and 3 units up (in quadrant II).
Also called the Cartesian coordinate system, named after René Descartes (1596–1650).

🔗 Relations and their properties

🔗 What is a relation

Relation: Any set of ordered pairs.

In algebra, we focus on sets of ordered pairs (x, y) in the coordinate plane.
Coordinates are typically related by an algebraic equation.
Example: The equation y = |x| − 2 defines a relation; we can find ordered pairs like (−3, 1), (−2, 0), (0, −2), etc.

📊 Graphs of relations

Graph: A visual representation of a relation on a rectangular coordinate plane.

Plot the ordered pairs as points.
Any curve on the coordinate plane represents a set of ordered pairs and defines a relation.
Arrows on lines indicate the relation continues without bounds.

🎯 Domain and range

Domain: The set consisting of all of the first components of a relation (all x-values).

Range (or codomain): The set consisting of all second components of a relation (all y-values).

We can determine domain and range from a graph by identifying the extent of x-values and y-values.
Example: If a graph shows x-values from −8 onward, the domain is [−8, ∞); if y-values start at 0 and go up, the range is [0, ∞).

⚙️ Functions—special relations

⚙️ Definition of a function

Function: A relation where each element in the domain corresponds to exactly one element in the range.

Every x-value must correspond to exactly one y-value.
Example: {(−1, 4), (0, 7), (2, 3), (3, 3), (4, −2)} is a function because no x-value repeats.
Non-example: {(−4, −3), (−2, 6), (0, 3), (3, 5), (3, 7)} is not a function because x = 3 corresponds to both y = 5 and y = 7.

🔍 Identifying functions from ordered pairs

Check if any x-value appears more than once.
If an x-value repeats with different y-values, it is not a function.
If all x-values are unique (or repeat with the same y-value), it is a function.

📏 The vertical line test

Vertical line test: If any vertical line intersects the graph more than once, then the graph does not represent a function.

A vertical line represents one x-value.
The number of intersections shows how many y-values correspond to that x-value.
Example: The graph of y = |x| − 2 passes the test (function); the graph of x = |y| + 1 fails (not a function—one x-value can correspond to two y-values).
Don't confuse: A horizontal line test checks something different (one-to-one functions); the vertical line test checks if a relation is a function at all.

🔤 Function notation and evaluation

🔤 What is function notation

Function notation: The notation f(x) = y, which reads "f of x is equal to y."

f is the function name (could also be g, h, C, R, etc.).
f(x) denotes the output value in the range for input x in the domain.
y and f(x) are interchangeable—they mean the same thing.
Not multiplication: f(x) does not mean "f times x."

🧮 Evaluating functions

To find f(a), substitute a for x in the function formula and simplify.
The value inside the parentheses is called the argument of the function.
Example: If f(x) = 2x − 3, then f(−5) = 2(−5) − 3 = −10 − 3 = −13.

Best practice: Replace the variable with parentheses first, then substitute.

Example: Given g(x) = x², find g(−2), g(1/2), and g(x + h):

g(−2) = (−2)² = 4
g(1/2) = (1/2)² = 1/4
g(x + h) = (x + h)² = x² + 2xh + h²

⚠️ Common mistake

Important: In general, f(x + h) ≠ f(x) + f(h).

Example with g(x) = x²:

g(x + h) = (x + h)² = x² + 2xh + h²
g(x) + g(h) = x² + h²
These are not equal because of the 2xh term.

🔄 Finding inputs from outputs

If asked to find x where f(x) = a, set the function equal to a and solve for x.
Example: Given f(x) = 5x + 7, find x where f(x) = 27:
- 27 = 5x + 7
- 20 = 5x
- x = 4
Check: f(4) = 5(4) + 7 = 27 ✓

📈 Reading function values from graphs

To find f(a) from a graph: locate x = a on the x-axis, find the corresponding y-value on the graph.
To find x where f(x) = b: locate y = b on the y-axis, find the corresponding x-value(s) on the graph.
Example: If the graph shows a point at (−5, 2), then f(−5) = 2 and x = −5 where f(x) = 2.

📝 Key takeaways from the excerpt

Concept	Definition	How to identify
Relation	Any set of ordered pairs	List of pairs or graph
Function	Relation where each x maps to exactly one y	No repeated x-values; passes vertical line test
Domain	Set of all x-values (inputs)	All first coordinates; horizontal extent on graph
Range	Set of all y-values (outputs)	All second coordinates; vertical extent on graph
Function notation	f(x) = y	Interchangeable with y; not multiplication

Linear Functions and Their Graphs

🧭 Overview

🧠 One-sentence thesis

Linear functions can be graphed by plotting points or using slope and y-intercept, and their graphs provide a geometric way to solve linear equations and inequalities by finding where two lines intersect or comparing their positions.

📌 Key points (3–5)

Graphing by plotting points: solve for y, choose x-values, calculate corresponding y-values, plot the ordered pairs, and draw a line through them.
Slope measures steepness: slope is the ratio of vertical change (rise) to horizontal change (run), calculated as m = (y₂ − y₁)/(x₂ − x₁).
Slope-intercept form for quick graphing: y = mx + b, where m is the slope and (0, b) is the y-intercept; start at the y-intercept and mark off the slope to find a second point.
Common confusion—horizontal vs vertical lines: horizontal lines (y = k) have zero slope and are functions; vertical lines (x = k) have undefined slope and are NOT functions.
Graphical interpretation of equations: the solution to f(x) = g(x) is the x-value where the two graphs intersect; inequalities correspond to regions where one graph lies above or below the other.

📊 Plotting Points and Graphing Lines

📊 The plotting-points technique

The excerpt demonstrates graphing the equation 8x + 4y = 12 by first solving for y:

Rearrange to isolate y: y = −2x + 3
Identify the independent variable (x, the input) and dependent variable (y, the output whose value depends on x)
Choose several x-values (negative, zero, and positive)
Calculate the corresponding y-values
Plot the resulting ordered pairs and draw a straight line through them with arrows on both ends (indicating the line extends infinitely)

Example from the excerpt: for x = −2, y = −2(−2) + 3 = 7, giving the point (−2, 7); for x = 0, y = 3, giving (0, 3); and so on.

Plotting points: A way of determining a graph using a finite number of representative ordered pair solutions.

🔍 Why this works

A linear equation has infinitely many solutions
Any two points determine a unique line
Plotting several points ensures accuracy and confirms the line is straight

📐 Understanding Slope

📐 What slope measures

Slope: The incline of a line measured as the ratio of the vertical change to the horizontal change, often referred to as "rise over run."

The excerpt uses a real-world analogy: a 5% incline means for every 100 feet forward (horizontal), the height increases 5 feet (vertical).

📐 The slope formula

Given any two points (x₁, y₁) and (x₂, y₂):

Slope m = rise/run = (y₂ − y₁)/(x₂ − x₁) = Δy/Δx

Δ (Greek letter delta) represents "change in"
The order of subtraction must be consistent: subtract coordinates of the first point from the second in the same order for both numerator and denominator

Example from the excerpt: for points (−3, −5) and (2, 1), slope = (1 − (−5))/(2 − (−3)) = 6/5. Switching the order gives (−5 − 1)/(−3 − 2) = (−6)/(−5) = 6/5, the same result.

📐 Four geometric cases for slope

Line type	Slope value	Visual characteristic
Upward incline (left to right)	Positive	Rising line
Downward incline (left to right)	Negative	Falling line
Horizontal	Zero	Flat line (y = k)
Vertical	Undefined	Straight up-and-down (x = k)

Why horizontal lines have zero slope: All points share the same y-value, so rise = 0, making m = 0/run = 0.

Why vertical lines have undefined slope: All points share the same x-value, so run = 0, making m = rise/0, which is undefined (division by zero).

🎯 Slope-Intercept Form and Linear Functions

🎯 Converting to slope-intercept form

Slope-intercept form: Any nonvertical line can be written in the form y = mx + b, where m is the slope and (0, b) is the y-intercept.

The excerpt shows converting 3x − 4y = 8 (standard form) to y = (3/4)x − 2 (slope-intercept form) by solving for y.

🎯 The y-intercept

y-intercept: The point (or points) where a graph intersects the y-axis, expressed as an ordered pair (0, y).

At the y-intercept, x = 0
In y = mx + b, when x = 0, y = b
So the y-intercept is always (0, b)

🎯 Quick graphing method

To graph y = (3/4)x − 2:

Start at the y-intercept (0, −2)
From that point, mark off the slope: rise 3, run 4 (or down 3, left 4 for negative direction)
Draw a line through these two points

🎯 Linear functions

Linear function: Any function that can be written in the form f(x) = mx + b, where the slope m and b represent any real numbers.

Because y = f(x), ordered pairs can be written as (x, y) or (x, f(x))
The y-intercept is (0, f(0)) = (0, b)
Domain and range are both all real numbers (for nonvertical lines)
The vertical line test confirms these are functions

🎯 Finding the x-intercept

x-intercept: The point (or points) where a graph intersects the x-axis, expressed as an ordered pair (x, 0).

To find the x-intercept, set f(x) = 0 and solve for x.

Example from the excerpt: for f(x) = −(5/3)x + 6, set 0 = −(5/3)x + 6, solve to get x = 18/5, so the x-intercept is (18/5, 0).

🔲 Special Cases: Horizontal and Vertical Lines

🔲 Horizontal lines (constant functions)

Equation form: y = k or f(x) = c (where k and c are constants)
Slope: zero
x-intercept: none (unless the line is y = 0)
y-intercept: (0, k)
Domain: all real numbers
Range: {k} (a single value)
Is a function: Yes (passes vertical line test)

Example from the excerpt: g(x) = −2 is a horizontal line through (0, −2).

🔲 Vertical lines

Equation form: x = k (where k is a constant)
Slope: undefined
x-intercept: (k, 0)
y-intercept: none (unless the line is x = 0)
Domain: {k} (a single value)
Range: all real numbers
Is NOT a function: Fails the vertical line test because one x-value corresponds to infinitely many y-values

Don't confuse: Horizontal lines ARE functions (zero slope); vertical lines are NOT functions (undefined slope).

🔍 Graphical Interpretation of Equations and Inequalities

🔍 Solving f(x) = g(x) graphically

The solution to f(x) = g(x) is the x-value where the two graphs intersect.

Example from the excerpt: to solve (1/2)x + 1 = 3:

Graph f(x) = (1/2)x + 1 (a line with slope 1/2 and y-intercept (0, 1))
Graph g(x) = 3 (a horizontal line)
The graphs intersect at x = 4
Therefore, (1/2)x + 1 = 3 when x = 4

This geometric approach confirms the algebraic solution.

🔍 Solving f(x) ≥ g(x) graphically

The solution to f(x) ≥ g(x) consists of all x-values where the graph of f lies on or above the graph of g.

Example from the excerpt: to solve (1/2)x + 1 ≥ 3:

Graph both functions as before
Identify where f(x) is at or above g(x)
From the graph, this occurs when x ≥ 4
In interval notation: [4, ∞)

🔍 Solving f(x) < g(x) graphically

The solution consists of all x-values where the graph of f lies strictly below the graph of g.

Why this matters: Graphing provides a visual check for algebraic solutions and helps understand solution sets for inequalities as regions rather than just isolated points.

Modeling Linear Functions

🧭 Overview

🧠 One-sentence thesis

Linear functions can be constructed from geometric or real-world data by determining slope and y-intercept, then used to model applications like cost, revenue, and depreciation.

📌 Key points (3–5)

Finding equations from points: Given two points or a point and slope, you can write the equation using point-slope form or slope-intercept form.
Parallel vs perpendicular lines: Parallel lines share the same slope (m₁ = m₂); perpendicular lines have opposite reciprocal slopes (m₁ · m₂ = -1).
Mathematical modeling: Real-world data (costs, revenue, depreciation) can be expressed as linear functions to make predictions.
Common confusion: Interpolation (estimating within given data) vs extrapolation (predicting beyond given data)—both use the same function but extrapolation extends past known values.
Business applications: Profit equals revenue minus cost; the breakeven point occurs when profit equals zero.

📐 Constructing equations of lines

📐 Point-slope form

Point-slope form: y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is any point on the line.

This form is derived by applying the slope formula with a given point and a variable point (x, y).
Particularly useful when you know the slope and one point.
After substituting, solve for y to convert to slope-intercept form if needed.

📊 Finding slope from two points

Use the slope formula: m = (y₂ - y₁) / (x₂ - x₁)
Once you have the slope, substitute either point into point-slope form.
Example: Given (-3, 6) and (5, -4), the slope is m = (-4 - 6)/(5 - (-3)) = -10/8 = -5/4.

🎯 Using slope-intercept form

If you can identify the y-intercept (0, b) directly, use f(x) = mx + b.
Substitute a known point to solve for b if it's not immediately visible.
Example: With slope m = -5/4 and point (-3, 6), substitute to find b = 9/4.

🔀 Parallel and perpendicular lines

↔️ Parallel lines

Two nonvertical lines are parallel if their slopes are equal: m₁ = m₂.
They never intersect and remain the same distance apart.
To find a parallel line through a given point: use the same slope as the original line, then apply point-slope form with the new point.
Example: A line parallel to y = (1/2)x + 1 through (3, -2) has slope m = 1/2 and equation y = (1/2)x - 7/2.

⊥ Perpendicular lines

Two nonvertical lines are perpendicular if the product of their slopes is -1: m₁ · m₂ = -1.
Their slopes are opposite reciprocals (also called negative reciprocals).
If m = a/b, then m⊥ = -b/a.
Example: If the original slope is -1/4, the perpendicular slope is +4/1 = 4.
Don't confuse: "opposite reciprocal" means flip the fraction AND change the sign.

🏗️ Mathematical modeling with linear functions

🏗️ What is mathematical modeling

Mathematical modeling: Using data to find equations that describe or model real-world applications.

Identify the dependent variable (what depends on something else) and independent variable (what you control or measure).
Find two ordered pairs from the given situation.
Use these pairs to calculate slope and y-intercept, then write the function.

💰 Cost and revenue functions

Cost function C(n): models the total cost of producing n units.
- Typically includes a fixed cost (setup fee) plus variable cost per unit.
- Example: C(n) = 62n + 5,280 means $5,280 fixed cost plus $62 per item.
Revenue function R(n): models income from selling n units.
- Usually price per unit times number of units.
- Example: R(n) = 150n means $150 per item sold.

📈 Profit and breakeven

📈 Profit function

Profit function: P(n) = R(n) - C(n), modeling profit as revenue less cost.

Profit can be positive (gain), negative (loss), or zero (breakeven).
Example: If R(n) = 150n and C(n) = 62n + 5,280, then P(n) = 88n - 5,280.

⚖️ Breakeven point

Breakeven point: The point at which profit is neither negative nor positive; profit equals zero.

Set P(n) = 0 and solve for n.
Example: 0 = 88n - 5,280 gives n = 60 units to break even.
Below this point, the business operates at a loss; above it, at a profit.

📉 Depreciation and growth models

📉 Linear depreciation

Linear depreciation model: A linear function used to describe the declining value of an item over time.

Typically uses time as the independent variable.
Slope is negative, representing value loss per time period.
Example: Equipment purchased for $12,000 and worth $9,000 after 4 years has slope m = (9,000 - 12,000)/(4 - 0) = -750.
The model V(t) = -750t + 12,000 shows value declining by $750 per year.

🔍 Interpolation vs extrapolation

Interpolation: Estimating values between given data points.
- Example: Using V(t) = -750t + 12,000 to find value at t = 2 years (between 0 and 4).
Extrapolation: Predicting values beyond given data points.
- Example: Using the same function to predict value at t = 10 years (beyond the 4-year data).
Don't confuse: Both use the same function, but extrapolation is less reliable because it assumes the pattern continues outside the known range.

📊 Growth models

Linear growth uses a positive slope to model increasing values over time.
Example: Population or user registration growing at a constant rate per year.
Same construction method: find two data points, calculate slope, write the function.

Graphing the Basic Functions

🧭 Overview

🧠 One-sentence thesis

Seven basic functions—constant, identity, squaring, cubing, absolute value, square root, and reciprocal—form the foundation for graphing and understanding more complex functions, and their shapes, domains, and ranges must be memorized for use throughout the course.

📌 Key points (3–5)

Seven basic functions: constant, identity, squaring, cubing, absolute value, square root, and reciprocal functions each have distinct shapes and properties.
Polynomial vs nonpolynomial: the first four are polynomial functions; the last three are nonpolynomial.
Domain and range vary: some functions are defined for all real numbers, while others have restrictions (e.g., square root requires non-negative inputs; reciprocal excludes zero).
Piecewise functions: functions whose definition changes depending on the input value; evaluation requires selecting the correct piece based on the domain condition.
Common confusion: the greatest integer function assigns the greatest integer less than or equal to a value, not the nearest integer (it is not rounding).

📐 Polynomial basic functions

📏 Constant function

Constant function: any function of the form f(x) = c, where c is any real number.

Can be written as f(x) = 0x + c, showing slope = 0 and y-intercept (0, c).
Evaluating any x-value always returns c.
Graph: horizontal line.
Domain: all real numbers ℝ.
Range: the single value {c}.

🔷 Identity function

Identity function: the linear function defined by f(x) = x.

Evaluating any x returns that same value: f(0) = 0, f(2) = 2.
Written as f(x) = 1x + 0, with slope m = 1 and y-intercept (0, 0).
Graph: straight line through the origin at 45°.
Domain and range: all real numbers ℝ.

📈 Squaring function

Squaring function: the quadratic function defined by f(x) = x².

Squaring values in the domain: f(2) = 4, f(−2) = 4.
Squaring nonzero values always produces positive results.
Graph: a curved shape called a parabola.
Domain: all real numbers ℝ.
Range: all y-values ≥ 0, written [0, ∞).

📊 Cubing function

Cubing function: the cubic function defined by f(x) = x³.

Raises domain values to the third power.
Results can be positive, zero, or negative: f(1) = 1, f(0) = 0, f(−1) = −1.
Domain and range: all real numbers ℝ.

🔀 Nonpolynomial basic functions

📏 Absolute value function

Absolute value function: the function defined by f(x) = |x|.

Output represents the distance to the origin on a number line.
Evaluating any nonzero x always yields a positive result: f(−2) = 2, f(2) = 2.
Domain: all real numbers ℝ.
Range: all y-values ≥ 0, written [0, ∞).
Can be written as a piecewise function:
- f(x) = x if x ≥ 0
- f(x) = −x if x < 0

√ Square root function

Square root function: the function defined by f(x) = √x.

Not defined as a real number if x-values are negative.
Smallest domain value is zero: f(0) = 0, f(4) = 2.
Domain: real numbers ≥ 0, written [0, ∞).
Range: real numbers ≥ 0, written [0, ∞).

➗ Reciprocal function

Reciprocal function: the function defined by f(x) = 1/x.

A rational function with one restriction: x ≠ 0.
Vertical asymptote at the y-axis: as x-values approach zero, reciprocals tend toward ±∞.
- Example: f(1/10) = 10, f(1/100) = 100, f(1/1000) = 1000.
Horizontal asymptote at the x-axis: as x-values become very large, y-values tend toward zero.
- Example: f(10) = 0.1, f(100) = 0.01, f(1000) = 0.001.
Domain and range: all real numbers except 0, written (−∞, 0) ∪ (0, ∞).

Don't confuse: A vertical asymptote is a vertical line the graph approaches infinitely closely; a horizontal asymptote is a horizontal line approached as x-values tend toward ±∞.

🧩 Piecewise functions

🧩 Definition and structure

Piecewise function (or split function): a function whose definition changes depending on the value in the domain.

The definition used depends on which interval or condition the input satisfies.
Example: the absolute value function can be written piecewise:
- f(x) = x if x ≥ 0
- f(x) = −x if x < 0
Each "piece" is graphed over its specified domain interval on the same coordinate plane.

📊 Graphing piecewise functions

Graph each piece separately over its domain interval.
Use open dots for endpoints not included (strict inequality < or >).
Use closed dots for endpoints included (inequality ≤ or ≥).
Example: g(x) = x² if x < 0; √x if x ≥ 0
- Graph the squaring function for negative x with an open dot at the origin.
- Graph the square root function for non-negative x with a closed dot at the origin.

🔢 Evaluating piecewise functions

The value in the domain determines which definition to use.
Check which condition the input satisfies, then apply the corresponding formula.
Example: if h(t) = 7t + 3 for t < 0 and −16t² + 32t for t ≥ 0:
- h(−5) uses the first piece: 7(−5) + 3 = −32.
- h(0) uses the second piece: −16(0)² + 32(0) = 0.
- h(3) uses the second piece: −16(3)² + 32(3) = −48.

🔢 Greatest integer function

🔢 Definition and behavior

Greatest integer function: the function that assigns any real number x to the greatest integer less than or equal to x, denoted f(x) = [[x]].

Also called the floor function.
Examples:
- f(2.7) = 2
- f(π) = 3
- f(0.23) = 0
- f(−3.5) = −4 (the greatest integer ≤ −3.5 is −4, not −3)

Don't confuse: This is not rounding. It always takes the greatest integer less than or equal to the input, which for negative numbers means moving further from zero.

📉 Graph and applications

The graph consists of horizontal line segments (steps).
For each interval [n, n+1), the function value is the integer n.
Domain: all real numbers ℝ.
Range: the set of integers ℤ.
Has many applications in computer science.

🎯 Key takeaways

Function type	Examples	Key property
Polynomial	constant, identity, squaring, cubing	Smooth curves or lines; defined for all real numbers
Nonpolynomial	absolute value, square root, reciprocal	May have restrictions on domain; absolute value and square root have restricted range
Piecewise	Split definitions	Definition changes based on input value; use correct piece for evaluation
Greatest integer	Floor function	Assigns greatest integer ≤ x; creates step graph

Memorization required: the shape, domain, and range of each basic function should be committed to memory for use throughout the course.
Graphing method: plot points to determine the general shape of each function.

Using Transformations to Graph Functions

🧭 Overview

🧠 One-sentence thesis

Transformations allow us to sketch graphs of complex functions by systematically shifting, reflecting, or stretching a basic parent function according to algebraic changes in the function's formula.

📌 Key points (3–5)

Two types of transformations: rigid transformations (translations and reflections) preserve shape and size but change location; non-rigid transformations (dilations) change size or shape.
Vertical vs horizontal shifts: adding/subtracting outside the function moves the graph up/down; adding/subtracting inside (before applying the function) moves it left/right—the horizontal direction is opposite to what you might expect.
Reflections: multiplying the output by −1 reflects across the x-axis; multiplying the input by −1 reflects across the y-axis.
Common confusion: horizontal shifts are counterintuitive—f(x + h) shifts left h units, while f(x − h) shifts right h units.
Dilations: multiplying the entire function by a constant a stretches vertically if |a| > 1, or horizontally if 0 < |a| < 1; negative values also produce reflections.

📐 Rigid transformations: Translations

📍 Vertical translations

Vertical translation: a rigid transformation that shifts a graph up or down relative to the original graph.

Achieved by adding or subtracting a constant after applying the function.
Shift up k units: F(x) = f(x) + k (where k is positive).
Shift down k units: F(x) = f(x) − k (where k is positive).
Example: if f(x) = x², then g(x) = x² − 3 shifts the parabola down 3 units, and h(x) = x² + 3 shifts it up 3 units.
The shape and size remain unchanged; only the vertical position changes.

📍 Horizontal translations

Horizontal translation: a rigid transformation that shifts a graph left or right relative to the original graph.

Achieved by adding or subtracting a constant inside the function (to the x-coordinate before applying the function).
Shift left h units: F(x) = f(x + h) (where h is positive).
Shift right h units: F(x) = f(x − h) (where h is positive).
Don't confuse: the direction is opposite to the sign—adding inside shifts left, subtracting inside shifts right.
Example: g(x) = (x + 3)² shifts the parabola 3 units to the left; h(x) = (x − 3)² shifts it 3 units to the right.

🔄 Combining translations

Translations can be applied in any order without affecting the final graph.
Example: g(x) = |x + 3| − 5 involves a horizontal shift left 3 units and a vertical shift down 5 units.
When graphing, start with the basic function, apply the horizontal shift, then the vertical shift (or vice versa).

🪞 Rigid transformations: Reflections

🪞 What reflections do

Reflection: a transformation in which a mirror image of the graph is produced about an axis.

Reflections are rigid transformations—they change orientation but not shape or size.
Two types: reflection about the x-axis and reflection about the y-axis.

🪞 Reflection about the x-axis

Formula: F(x) = −f(x).
Multiply each y-coordinate by −1.
The graph flips upside down across the horizontal axis.
Example: if f(x) = √x, then h(x) = −√x reflects the square root curve below the x-axis.

🪞 Reflection about the y-axis

Formula: F(x) = f(−x).
Multiply each x-coordinate by −1 before applying the function.
The graph flips left-to-right across the vertical axis.
Example: if f(x) = √x, then g(x) = √(−x) reflects the curve across the y-axis.

🪞 Order of operations with reflections

When multiple transformations are present, apply reflections first, then translations.
Example: g(x) = −(x + 5)² + 3 involves reflection about the x-axis, horizontal shift left 5 units, and vertical shift up 3 units—start with the reflection.

🔍 Non-rigid transformations: Dilations

🔍 What dilations do

Dilation: a non-rigid transformation, produced by multiplying functions by a nonzero real number, which appears to stretch the graph either vertically or horizontally.

Unlike rigid transformations, dilations change the size or shape of the graph.
Formula: F(x) = a · f(x), where a is a nonzero real number.

🔍 Vertical stretches and compressions

If |a| > 1, the graph stretches vertically (appears steeper).
If 0 < |a| < 1, the graph compresses vertically (appears flatter, which looks like a horizontal stretch).
Example: g(x) = 4x² makes the parabola steeper; h(x) = (1/4)x² makes it flatter.

🔍 Combining dilations with reflections

If a is negative, the dilation also produces a reflection about the x-axis.
Example: g(x) = −2|x − 5| − 3 involves a vertical stretch by factor 2, reflection about the x-axis, horizontal shift right 5 units, and vertical shift down 3 units.
When graphing, handle the dilation and reflection together first, then apply translations.

📋 Summary table of transformations

Transformation	Formula	Effect	Type
Vertical shift up	F(x) = f(x) + k	Move graph up k units	Rigid
Vertical shift down	F(x) = f(x) − k	Move graph down k units	Rigid
Horizontal shift left	F(x) = f(x + h)	Move graph left h units	Rigid
Horizontal shift right	F(x) = f(x − h)	Move graph right h units	Rigid
Reflection about x-axis	F(x) = −f(x)	Flip upside down	Rigid
Reflection about y-axis	F(x) = f(−x)	Flip left-to-right	Rigid
Dilation	F(x) = a · f(x)	Stretch/compress vertically or horizontally	Non-rigid

📋 Key takeaway for graphing

Identify the basic parent function first.
Apply transformations in order: reflections and dilations first, then translations.
The order of horizontal and vertical translations does not matter.
Understanding transformations geometrically often leads to more elegant solutions in mathematics.

Solving Absolute Value Equations and Inequalities

🧭 Overview

🧠 One-sentence thesis

Absolute value equations and inequalities can be solved by isolating the absolute value and then applying specific theorems that convert them into standard linear equations or compound inequalities based on whether the problem involves equals, less-than, or greater-than relations.

📌 Key points (3–5)

What absolute value represents: the distance from zero on a number line, always positive or zero.
Three solution patterns: equations (|X| = p) split into two cases; "less than" inequalities (|X| ≤ p) become bounded intervals; "greater than" inequalities (|X| ≥ p) become two unbounded intervals.
Isolate first: before applying any theorem, the absolute value expression must be isolated on one side.
Common confusion: absolute value equal to a negative number has no solution; absolute value greater than a negative number is satisfied by all real numbers.
Special cases: some absolute value equations have only one solution (when the result equals zero) or no solution (when isolated absolute value equals a negative).

📐 Understanding absolute value

📏 Definition and meaning

Absolute value of a real number a, denoted |a|, is the distance between zero and the graph of that number on the number line.

Distance is always non-negative: |3| = 3 and |−3| = 3.
Algebraically, absolute value is a piecewise function: |a| = a if a ≥ 0, and |a| = −a if a < 0.
The expression inside the absolute value bars is called the argument.

🔢 Why two solutions often appear

If |x| = 3, then x can be 3 or −3 (both are distance 3 from zero).
In general, |X| = p (where p > 0) means X = p or X = −p.
The argument can be positive or negative to produce the same absolute value.

🔧 Solving absolute value equations

🎯 The basic theorem

If |X| = p, then X = −p or X = p.

This applies only when p is positive.
The argument X must equal either the positive or negative version of p.

📝 Step-by-step process

Isolate the absolute value: rearrange so |expression| stands alone on one side.
Set up two equations: argument = p and argument = −p.
Solve each equation: treat them as separate linear equations.
Check solutions: substitute back into the original equation to verify.

Example: To solve 2|5x − 1| − 3 = 9, first add 3 to both sides (2|5x − 1| = 12), then divide by 2 (|5x − 1| = 6), then split into 5x − 1 = 6 or 5x − 1 = −6, giving x = 7/5 or x = −1.

⚠️ Special equation cases

One solution: |X| = 0 means X = 0 only (zero distance from origin).
No solution: |X| = negative number is impossible (distance cannot be negative).

Example: |x + 7| + 5 = 4 isolates to |x + 7| = −1, which has no solution.

🔄 Two absolute values

If |X| = |Y|, then X = −Y or X = Y.

When two absolute values are equal, their arguments are either the same or opposites.
Set the arguments equal to each other and also equal to each other's opposite.

Example: |2x − 5| = |x − 4| becomes 2x − 5 = x − 4 or 2x − 5 = −(x − 4), yielding x = 1 or x = 3.

📉 Solving "less than" inequalities

📊 The bounded interval theorem

If |X| ≤ p, then −p ≤ X ≤ p.

"Less than" absolute value inequalities describe all numbers within distance p from zero.
The solution is a single bounded interval.
Works for both ≤ (inclusive) and < (strict).

🔍 Solution process

Isolate the absolute value.
Rewrite as a compound inequality with the argument bounded between −p and p.
Solve the compound inequality.
Express in interval notation; graph with closed dots for ≤ or open dots for <.

Example: |x + 2| < 3 becomes −3 < x + 2 < 3, then −5 < x < 1, written as (−5, 1).

🎨 Graphical interpretation

Let f(x) = |argument| and g(x) = p.
The solution consists of all x-values where the graph of f is below the graph of g.
The absolute value function forms a V-shape; the horizontal line at height p intersects it at two points.

📈 Solving "greater than" inequalities

📊 The unbounded intervals theorem

If |X| ≥ p, then X ≤ −p or X ≥ p.

"Greater than" absolute value inequalities describe numbers whose distance from zero exceeds p.
The solution is two unbounded intervals (one on each side).
Works for both ≥ (inclusive) and > (strict).

🔍 Solution process

Isolate the absolute value.
Rewrite as a compound inequality: argument ≤ −p OR argument ≥ p.
Solve each inequality separately.
Express using union notation: (−∞, −p] ∪ [p, ∞).

Example: |x + 2| > 3 becomes x + 2 < −3 or x + 2 > 3, giving x < −5 or x > 1, written as (−∞, −5) ∪ (1, ∞).

🎨 Graphical interpretation

The solution consists of all x-values where the graph of f(x) = |argument| is above g(x) = p.
The V-shaped absolute value function rises above the horizontal line in two separate regions.

⚠️ Special inequality cases

All real numbers: |X| > negative number is always true (absolute value is always non-negative).
No solution: |X| ≤ negative number is impossible.

Example: |x + 1| + 4 ≤ 3 isolates to |x + 1| ≤ −1, which has no solution (Ø).

🗂️ Summary of three cases

Theorem	Solution type	Example
X	= p	X = −p or X = p
X	≤ p	−p ≤ X ≤ p
X	≥ p	X ≤ −p or X ≥ p

Don't confuse: The key distinction is the relation symbol (=, <, >) which determines which theorem to apply after isolating the absolute value.

Solving Inequalities with Two Variables

🧭 Overview

🧠 One-sentence thesis

Inequalities with two variables have solution sets that form shaded regions on a coordinate plane, bounded by lines or curves that may or may not be included depending on whether the inequality is strict or inclusive.

📌 Key points (3–5)

What the solution set looks like: a shaded region (half-plane for linear inequalities) rather than a single line or set of points.
Boundary inclusion rule: strict inequalities (< or >) use dashed boundaries; inclusive inequalities (≤ or ≥) use solid boundaries.
Test point method: choose any point not on the boundary, substitute it into the inequality, and shade the region containing it if true (or the opposite region if false).
Common confusion: "greater than" does not always mean "shade above"—the rule only applies when the inequality is in slope-intercept form (y > mx + b); always use a test point to verify.
Extension to nonlinear: the same graphing process applies to parabolic, absolute value, and other nonlinear boundaries.

📐 Understanding solution sets

📐 What a solution set is

Linear inequality with two variables: an inequality relating linear expressions with two variables; the solution set is a region defining half of the plane.

Unlike linear equations (which graph as lines), linear inequalities graph as regions.
Every ordered pair in the shaded region satisfies the inequality.
Example: for y ≤ (3/2)x + 3, points like (0,0), (2,1), and (−2,−1) all satisfy the inequality, while (−2,3) does not.

🔍 Boundary vs. interior

The boundary is the line (or curve) you get by replacing the inequality symbol with an equals sign.
Points on the boundary satisfy the equation but may or may not satisfy the inequality.
The "or equal to" part of ≤ or ≥ determines whether boundary points are solutions.

🖊️ Graphing technique

🖊️ Step 1: Graph the boundary

Replace the inequality symbol with = to get the boundary equation.
Dashed line/curve: use for strict inequalities (< or >); boundary points are not solutions.
Solid line/curve: use for inclusive inequalities (≤ or ≥); boundary points are solutions.
Example: y < (3/2)x + 3 has a dashed boundary; y ≤ (3/2)x + 3 has a solid boundary.

🧪 Step 2: Use a test point

Test point: a point not on the boundary used to determine which half-plane contains the solutions.

The origin (0,0) is a common test point (unless it lies on the boundary).
Substitute the test point's coordinates into the inequality.
If the result is true, shade the region containing the test point.
If the result is false, shade the opposite region.
Example: testing (0,0) in y ≥ (3/2)x + 3 gives 0 ≥ 3, which is false, so shade the region that does not contain (0,0).

🎨 Step 3: Shade the solution region

Shade the half-plane (or region) that contains all solutions.
The excerpt emphasizes: verify by testing additional points inside and outside the shaded region.

⚠️ Common pitfalls

⚠️ "Greater than" does not always mean "shade above"

The rule "y > mx + b means shade above" only works when the inequality is already solved for y in slope-intercept form.
If the boundary is in standard form (e.g., 2x − 5y ≥ −10), this shortcut does not apply.
Don't confuse: the excerpt shows an example where 2x − 5y ≥ −10 requires shading below the boundary, even though the symbol is ≥.
Best practice: always use a test point rather than relying on the inequality symbol alone.

⚠️ Reversing the inequality when solving for y

When you divide or multiply both sides by a negative number, you must reverse the inequality symbol.
Example: starting from 2x − 5y ≥ −10, subtracting 2x and dividing by −5 gives y ≤ (2/5)x + 2 (the ≥ becomes ≤).

🌐 Nonlinear inequalities

🌐 Same process, different boundaries

The graphing steps are identical for nonlinear inequalities.
The boundary can be a parabola, absolute value graph, cubic, or other curve.
Example: y > x² has a parabolic boundary (dashed), and the region above the parabola is shaded.

🌐 Parabolic boundaries

Inequality form	Boundary shape	Typical shading
y ≤ (x − 1)² − 2	Parabola (solid)	Inside/below the parabola
y ≥ (x − 1)² − 2	Parabola (solid)	Outside/above the parabola

The excerpt provides examples with shifted parabolas: y ≤ (x + 2)² − 1 and y ≥ x² + 3.
Always test a point to confirm which region to shade.

🌐 Absolute value and other functions

Absolute value inequalities (e.g., y < |x − 1| − 3) graph with V-shaped boundaries.
The same three-step process applies: graph the boundary (dashed or solid), test a point, and shade the appropriate region.

Linear Systems with Two Variables and Their Solutions

🧭 Overview

🧠 One-sentence thesis

A linear system with two variables has a solution that corresponds geometrically to the point where two lines intersect, and understanding whether lines intersect, coincide, or are parallel determines whether the system has one solution, infinitely many solutions, or no solution.

📌 Key points (3–5)

What a solution is: an ordered pair (x, y) that satisfies both equations simultaneously and corresponds to a point of intersection on a graph.
The graphing method: solve by graphing both lines on the same axes and finding where they cross.
Three types of systems: independent (one solution), dependent (infinitely many solutions—same line), and inconsistent (no solution—parallel lines).
Common confusion: dependent vs inconsistent—dependent systems have the same slope and same y-intercept (same line), while inconsistent systems have the same slope but different y-intercepts (parallel lines).
Accuracy limitation: graphing is not ideal when solutions involve non-integer coordinates; algebraic methods are more precise.

📐 What is a linear system and its solution

📐 Definition of a linear system

System of equations: a set of two or more equations with the same variables.

Linear system: a set of two or more linear equations with the same variables.

In this section, the focus is on systems with exactly two linear equations and two variables.
Example format: {2x − 3y = 0; −4x + 2y = −8}

✅ What counts as a solution

Solution to a linear system (or simultaneous solution): an ordered pair (x, y) that solves both equations.

To check a solution, substitute the x- and y-values into each equation.
Both equations must produce true statements.
Example: (3, 2) is a solution if substituting x = 3 and y = 2 into both equations yields true statements like 0 = 0 and −8 = −8.

Don't confuse: A solution must satisfy both equations, not just one. If an ordered pair works in one equation but not the other, it is not a solution to the system.

🖼️ The graphing method

🖼️ How the graphing method works

Geometric interpretation: each linear equation represents a line; a solution is the point where the two lines intersect.
Steps:
1. Rewrite each equation in slope-intercept form (y = mx + b) to create an equivalent system.
2. Graph both lines on the same set of axes.
3. Identify the intersection point.
4. Check the point by substituting into the original equations.

Equivalent system: a system consisting of equivalent equations that share the same solution set.

Example: The system {2x − 3y = 0; −4x + 2y = −8} can be rewritten as {y = (2/3)x; y = 2x − 4}, which is easier to graph.

⚠️ Limitations of graphing

Graphing is not accurate when the solution has non-integer coordinates.
The excerpt emphasizes: "take care to choose a good scale and use a straightedge... accuracy is very important."
Algebraic methods (covered in later sections) are more precise.
Always check your graphical solution by substituting into the original equations.

🔀 Three types of systems

🔀 Consistent vs inconsistent

Consistent system: a system with at least one solution.

Inconsistent system: a system with no simultaneous solution (denoted Ø).

Inconsistent systems graph as parallel lines that never intersect.
These lines have the same slope but different y-intercepts.

🔁 Independent systems (one solution)

Independent system: a linear system with two variables that has exactly one ordered pair solution.

The two lines have different slopes, so they intersect at exactly one point.
The solution is written as a single ordered pair, e.g., (3, 2).
This is the most common case in the examples.

♾️ Dependent systems (infinitely many solutions)

Dependent system: a linear system with two variables that consists of equivalent equations; it has infinitely many ordered pair solutions.

The two equations represent the same line (same slope and same y-intercept).
When graphed, the lines coincide completely.
Solution notation: (x, mx + b), where x is any real number.
- This is shorthand for "all points on the line y = mx + b."
- Alternative notation: {(x, y) | y = mx + b} (set notation).

Example from the excerpt: The system {−2x + 3y = −9; 4x − 6y = 18} simplifies to {y = (2/3)x − 3; y = (2/3)x − 3}, which is the same line, so the solution is (x, (2/3)x − 3).

🆚 How to distinguish the three types

Type	Slopes	y-intercepts	Graph behavior	Solution
Independent	Different	Any	Lines intersect at one point	One ordered pair (x, y)
Dependent	Same	Same	Lines coincide (same line)	Infinitely many: (x, mx + b)
Inconsistent	Same	Different	Lines are parallel	No solution: Ø

Key insight: When you rewrite both equations in slope-intercept form, compare the slopes and y-intercepts to predict the type of system before graphing.

🎯 Key takeaways from the excerpt

🎯 Geometry and algebra connection

Solutions correspond to points where graphs intersect.
Understanding the geometric meaning helps interpret algebraic results.

🎯 Checking solutions

Always verify by substituting into both original equations.
A true statement (e.g., 6 = 6) confirms the solution works for that equation.

🎯 Notation for dependent systems

The excerpt uses the shortened form (x, mx + b) to express infinitely many solutions.
This means "x can be any real number, and y is determined by the equation y = mx + b."

Don't confuse: The notation (x, mx + b) does not mean a single point; it represents all points on the line.

Solving Linear Systems with Two Variables

🧭 Overview

🧠 One-sentence thesis

The substitution and elimination methods are two algebraic techniques for solving linear systems, each with distinct strengths depending on the coefficients and structure of the equations.

📌 Key points (3–5)

Substitution method: solve one equation for a variable, substitute into the other, then back substitute to find both coordinates.
Elimination method: multiply equations to create opposite coefficients for one variable, add the equations to eliminate that variable, then back substitute.
Common confusion: dependent systems yield identities (e.g., 2 = 2) and infinitely many solutions in the form (x, mx + b); inconsistent systems yield false statements (e.g., −6 = −16) and no solution.
Choosing a method: substitution works well when a variable has coefficient 1; elimination is better when coefficients are not 1 (avoids fractions).
Clearing fractions/decimals: multiply both sides by the LCD or appropriate power of 10 before applying elimination.

🔄 The substitution method

🔄 How substitution works

The substitution method solves one equation for one variable, then substitutes that expression into the other equation.

This leaves a single equation with one variable.
After solving for that variable, back substitute into the earlier expression to find the other coordinate.

Back substitute: once a value is found for a variable, substitute it back into one of the original equations (or its equivalent) to determine the corresponding value of the other variable.

📝 Step-by-step process

Solve one equation for one variable (choose the easiest).
Substitute the resulting expression into the other equation.
Solve the resulting single-variable equation.
Back substitute to find the second coordinate.
Write the solution as an ordered pair (x, y).

Example: Given 2x + y = −3 and 3x − 2y = −8, solve the first for y to get y = −2x − 3, substitute into the second equation, solve for x = −2, then back substitute to find y = 1. Solution: (−2, 1).

⚠️ When substitution is less ideal

If no variable has coefficient 1, solving for a variable introduces fractions.
Fractional coefficients are tedious to work with.
In such cases, elimination is usually preferable.

➕ The elimination method

➕ How elimination works

The elimination method uses the addition property of equations: if A = B and C = D, then A + C = B + D.

Multiply one or both equations by constants so that one variable has opposite coefficients in the two equations.
Add the equations together; the variable with opposite coefficients cancels out.
Solve the resulting single-variable equation, then back substitute.

🎯 Step-by-step process

Rewrite both equations in standard form (ax + by = c) if needed.
Identify which variable to eliminate (x or y).
Find the least common multiple (LCM) of the coefficients of that variable.
Multiply each equation by the appropriate factor to create opposite coefficients.
Add the equations to eliminate the variable.
Solve for the remaining variable.
Back substitute to find the other coordinate.

Example: Given 5x − 3y = −1 and 3x + 2y = 7, multiply the first by 2 and the second by 3 to get 10x − 6y = −2 and 9x + 6y = 21. Adding gives 19x = 19, so x = 1. Back substitute to find y = 2. Solution: (1, 2).

🧹 Clearing fractions and decimals

Fractions: multiply both sides by the LCD of all denominators before starting elimination.
Decimals: multiply both sides by 10, 100, etc., to convert to integers.
This simplifies arithmetic and reduces errors.

Example: For −(1/10)x + (1/2)y = 4/5, multiply by 10 to get −x + 5y = 8.

🔍 Special cases: dependent and inconsistent systems

🔍 Dependent systems (infinitely many solutions)

During solving, you reach an identity (always-true statement like 2 = 2 or 0 = 0).
The two equations represent the same line.
Solution set: (x, mx + b), where x is any real number.

Example: Solving −5x + y = −1 and 10x − 2y = 2 by substitution leads to 2 = 2. The solution is (x, 5x − 1).

❌ Inconsistent systems (no solution)

During solving, you reach a contradiction (false statement like −6 = −16).
The two lines are parallel and never intersect.
Solution: Ø (empty set).

Example: Solving −7x + 3y = 3 and 14x − 6y = −16 by substitution leads to −6 = −16. No solution exists.

🧭 How to recognize each case

Outcome during solving	System type	Solution
Identity (e.g., 0 = 0)	Dependent	(x, mx + b)
Contradiction (e.g., −6 = −16)	Inconsistent	Ø
Single value for variable	Independent	One ordered pair (x, y)

Don't confuse: an identity means the lines coincide (same line); a contradiction means the lines are parallel (different lines, same slope).

🎯 Choosing the right method

🎯 When to use substitution

A variable has coefficient 1 in one of the equations.
Easy to isolate a variable in one step.

Example: Given y = 5x + 15 and y = −5x + 5, substitution is natural because y is already isolated.

🎯 When to use elimination

No variable has coefficient 1, or isolating a variable would create fractions.
Both equations are in standard form or easily converted.

Example: Given 2x − 3y = 9 and 5x − 8y = −16, elimination avoids fractional coefficients.

📊 Summary of method strengths

Method	Strengths	Weaknesses
Graphing	Visual understanding; shows intersection	Inaccurate; impractical for non-integer solutions
Substitution	Fully algebraic; works well with coefficient 1	Often leads to fractions if no coefficient is 1
Elimination	Fully algebraic; avoids fractions	Does not generalize to nonlinear systems

🔑 Key takeaway

Both substitution and elimination are valid and produce the same correct result.
Choose based on the structure of the equations: coefficient of 1 → substitution; otherwise → elimination.
Always verify the solution by substituting back into both original equations.

Applications of Linear Systems with Two Variables

🧭 Overview

🧠 One-sentence thesis

Using two variables to set up linear systems simplifies word problems by allowing each unknown quantity to have its own variable, making relationships clearer and the algebraic setup more straightforward.

📌 Key points (3–5)

When to use two variables: Problems involving relationships between two unknowns become easier when each unknown gets its own variable, requiring two equations to solve.
Three main problem types: Relationship problems (integers, geometry), mixture problems (percentages and totals), and uniform motion problems (distance = rate × time).
Key distinction in mixtures: Percentage vs. amount—multiply the percentage by the total to get the actual amount of each component.
Common confusion in motion problems: With-current speed = base speed + current speed; against-current speed = base speed − current speed.
Setup strategy: Identify what you're asked to find, assign variables, then use the problem's constraints to build two independent equations.

🔢 Problems Involving Relationships Between Two Variables

🔢 Basic setup approach

When a problem describes relationships between two unknowns, assign a variable to each:

Let x represent one quantity
Let y represent the other quantity
Use the problem's sentences to build two separate equations

Why this helps: The excerpt emphasizes that "setting up word problems with two variables often simplifies the entire process, particularly when the relationships between the variables are not so clear."

🧮 Integer relationship problems

The excerpt shows an example where:

One sentence describes a sum: "4 times a larger integer and 5 times a smaller integer is 7"
Another sentence describes a difference: "twice the smaller integer is subtracted from 3 times the larger, the result is 11"

This naturally produces two equations:

4x + 5y = 7
3x − 2y = 11

Example: If one integer is 1 less than twice another and their sum is 20, you get two equations from those two facts, leading to the integers 7 and 13.

💰 Money and interest problems

Simple interest for one year: I = pr (principal × rate × 1 year)

The excerpt demonstrates a problem with $12,800 invested in two accounts at different rates (3.125% and 4.75%):

First equation: total principal → x + y = 12,800
Second equation: total interest → 0.03125x + 0.0475y = 465

Key point: Use decimal equivalents for percentages (3⅛% becomes 0.03125).

🪙 Coin problems

For a jar with nickels and dimes:

First equation counts coins: n + d = 58
Second equation counts value: 0.05n + 0.10d = 4.20

Don't confuse: The number of coins vs. the value of coins—multiply each coin count by its denomination to get value.

🧪 Mixture Problems

🧪 The percentage-amount distinction

The key distinction: multiply the percentage times the total to get the amount of each part of the mixture.

The excerpt gives a clear example:

A 20-ounce container with 2% saline solution contains:
- Salt: 0.02 × 20 = 0.4 ounces
- Water: 0.98 × 20 = 19.6 ounces

🧫 Setting up mixture equations

For combining solutions to reach a target concentration:

First equation: total amount → x + y = target total
Second equation: amount of active ingredient → (percentage₁)(x) + (percentage₂)(y) = (target percentage)(target total)

Example from excerpt: Mixing 1.8% and 3.2% saline solutions to get 35 ounces of 2.2% solution:

x + y = 35
0.018x + 0.032y = 0.022(35)

Result: 25 ounces of 1.8% solution and 10 ounces of 3.2% solution.

💧 Water as zero-percent solution

When mixing a concentrate with water:

Water contributes 0% of the active ingredient
The second equation simplifies because the water term drops out

Example: Mixing 80% antifreeze with water to get 48 liters at 25%:

x + y = 48
0.80x + 0(y) = 0.25(48), which simplifies to 0.80x = 12

🚗 Uniform Motion Problems

🚗 The fundamental formula

Distance = rate × time (D = r · t)

The excerpt emphasizes organizing data in a chart with columns for distance, rate, and time before setting up equations.

✈️ Multi-leg trip problems

When traveling by different modes:

Each leg gets its own row in the chart
Total time equation: x + y = total hours
Total distance equation: (rate₁)(x) + (rate₂)(y) = total miles

Example from excerpt: 4 hours total, 875 miles total, driving at 50 mph then flying at 320 mph:

x + y = 4
50x + 320y = 875

Result: 1½ hours driving, 2½ hours flying.

🌊 Current and wind problems

Key concept: Current or wind affects effective speed:

Situation	Effective speed
With current/wind	base speed + current speed
Against current/wind	base speed − current speed

Example from excerpt: Airplane traveling 240 miles in 2 hours with wind, then 135 miles in 1.5 hours against wind:

With wind: 240 = (x + w) · 2
Against wind: 135 = (x − w) · 1.5

Simplifying by dividing:

120 = x + w
90 = x − w

Result: Airplane speed 105 mph, wind speed 15 mph.

Don't confuse: The base speed of the vehicle vs. its effective speed when affected by current or wind—always use (base ± current) as the rate in the formula.

🎯 Identifying the unknown

The excerpt warns: "It is not always the case that time is the unknown quantity. Read the problem carefully and identify what you are asked to find; this defines your variables."

Sometimes you're asked to find speeds (as in the wind problem), sometimes times (as in the airport problem).

📋 Key Takeaways from the Excerpt

📋 When to use this approach

The excerpt states: "Use two variables as a means to simplify the algebraic setup of applications where the relationship between unknowns is unclear."

📋 Solution requirements

Read the problem several times
Remember: two variables require two equations
Answer in sentence form with correct units
The excerpt provides three footnoted problem categories: simple interest/money problems, mixture problems (percentages of totals), and uniform motion problems (distance-rate-time)

Budget: budget:token_budget1000000</budget:token_budget> Tokens used this turn: 1650 Remaining budget: 998350

Solving Linear Systems with Three Variables

🧭 Overview

🧠 One-sentence thesis

Linear systems with three variables can be solved systematically by using elimination to reduce the problem to two variables, then back-substituting to find all three coordinates, though some systems may have no solution or infinitely many solutions.

📌 Key points (3–5)

What a solution looks like: An ordered triple (x, y, z) that satisfies all three equations simultaneously.
The elimination method: Choose two pairs of equations, eliminate the same variable from each pair, solve the resulting two-variable system, then back-substitute.
Three possible outcomes: A single solution (consistent independent), no solution (inconsistent), or infinitely many solutions (consistent dependent).
Common confusion: Don't confuse the process with two-variable systems—you must eliminate the same variable twice using different equation pairs before reducing to a two-variable problem.
Geometric interpretation: Each equation represents a plane in three-dimensional space; solutions correspond to where planes intersect.

🔍 Checking solutions

✅ What makes a valid solution

Ordered triple: A set (x, y, z) that identifies position relative to the origin in three-dimensional space.

A solution must satisfy all three equations when you substitute the x, y, and z values.
If even one equation produces a false statement, the triple is not a solution.
The excerpt demonstrates checking by substituting (−2, 1, 3) into all three equations and verifying each produces a true statement.

🧮 The checking process

Substitute the x-value, y-value, and z-value into each equation in turn.
Simplify both sides of each equation.
Confirm that you get a true equality (e.g., −7 = −7) for all three equations.
Example: When checking whether (1, 4, 4/3) is a solution, the excerpt shows it fails the third equation, so it is not a solution.

🔧 The elimination method

📐 Step-by-step process

The excerpt outlines a four-step procedure:

First elimination: Choose any two equations and eliminate one variable.
Second elimination: Choose a different pair of equations and eliminate the same variable.
Solve the reduced system: You now have two equations with two variables; solve using familiar methods.
Back-substitute: Use the values you found to determine the third variable by substituting into one of the original equations.

🎯 Key technique details

You can choose which variable to eliminate first—the excerpt notes "it does not matter which variable we initially choose to eliminate."
Line up coefficients by multiplying equations by appropriate constants.
Add or subtract equations to cancel out the chosen variable.
Example: The excerpt shows eliminating z by multiplying the first equation by 3 and adding it to the second equation, then multiplying the first equation by −2 and adding it to the third equation.

🔄 Working with the reduced system

After two eliminations, you have a standard two-variable system.
Use elimination or substitution to find two of the three variables.
Once you have two variables, pick any original equation to find the third.
Don't confuse: You must use an original three-variable equation for back-substitution, not one of the intermediate two-variable equations.

🚫 Special cases: inconsistent and dependent systems

❌ Inconsistent systems (no solution)

What happens: During elimination, you arrive at a false statement like 0 = −42.
What it means: The system has no simultaneous solution; the three planes do not all intersect at a common point.
Geometric interpretation: The excerpt shows several configurations—planes may be parallel, or they may intersect in pairs but have no common point.

Example from the excerpt: When solving a system, the elimination process led to 15x + 18z = 48 and −15x − 18z = −25. Adding these gives 0 = −42, which is false, so the system is inconsistent.

♾️ Dependent systems (infinitely many solutions)

What happens: During elimination, you arrive at a true statement like 0 = 0.
What it means: The system has infinitely many solutions; the equations are not all independent.
How to express solutions: Write the solution as an ordered triple in terms of one variable (usually x).

Example from the excerpt: After elimination produces 5x − y = −10 and −5x + y = 10, adding gives 0 = 0. The solution is expressed as (x, 5x + 10, 13x + 25), meaning for any value of x, you can compute corresponding y and z values.

🗺️ Geometric configurations

The excerpt illustrates that infinitely many solutions can occur when:

Three planes intersect along a common line.
All three planes are actually the same plane.
Other configurations where the planes share infinitely many common points.

🌍 Applications with three unknowns

🎭 Setting up the system

Identify variables: Clearly define what each variable represents (e.g., x = number of adult tickets).
Write three equations: Look for three independent relationships in the problem.
Common relationships: Total quantities, total values, and comparative statements ("twice as many as").

📝 Example structure

The excerpt provides a theater ticket problem:

63 total tickets sold → equation (1): x + y + z = 63
Total sales of $444 with different prices → equation (2): 8x + 4y + 6z = 444
Twice as many adult tickets as child and senior combined → equation (3): x = 2(y + z), which rearranges to x − 2y − 2z = 0

The solution (42, 9, 12) means 42 adult tickets, 9 child tickets, and 12 senior tickets.

🔑 Key application strategies

Look for statements about totals (sums of quantities or values).
Look for comparative relationships (one quantity is twice another, etc.).
Make sure all three equations are independent—they should provide different information.
After solving, interpret the ordered triple in the context of the original problem.

Matrices and Gaussian Elimination

🧭 Overview

🧠 One-sentence thesis

Gaussian elimination transforms a linear system into upper triangular form using elementary row operations on an augmented matrix, enabling efficient solution by back substitution.

📌 Key points (3–5)

Upper triangular form: A system arranged so variables disappear in successive equations, making back substitution straightforward.
Matrix representation: Linear systems can be written as augmented matrices (coefficients plus constants) to streamline solving.
Elementary row operations: Three operations (swap rows, multiply a row, add multiples of rows) produce equivalent systems without changing solutions.
Common confusion: A row of all zeros means dependent (infinitely many solutions) if the constant is zero, but inconsistent (no solution) if the constant is nonzero.
Why it matters: Gaussian elimination is more efficient than elimination for larger systems and is implemented in calculators and computer algebra systems.

🔺 Upper triangular form and back substitution

🔺 What upper triangular form means

Upper triangular form: A linear system arranged so that variable x does not appear after the first equation and variable y does not appear after the second equation.

For two variables: the second equation contains only y.
For three variables: the second equation has no x, and the third equation has only z.
The coefficients form a triangular shape with zeros below the main diagonal.

🔙 How back substitution works

Start with the last equation (which has only one variable) and solve it.
Substitute that value into the previous equation to find the next variable.
Continue working backward until all variables are found.
Example: If the third equation gives z = -1, substitute into the second equation to find y, then substitute both into the first equation to find x.

Don't confuse: Back substitution only works when the system is already in upper triangular form—it is not a method for getting to that form.

🔢 Matrix representation

🔢 What a matrix is

Matrix: A rectangular array of numbers consisting of rows and columns.

Rows represent equations; columns represent variables.
Matrices organize coefficients without writing variables or operation symbols.

📊 Types of matrices for linear systems

Matrix type	What it contains	Example structure (3 variables)
Coefficient matrix	Only the coefficients of variables	Three columns (for x, y, z coefficients)
Augmented matrix	Coefficients plus the constants	Coefficient matrix with a dashed line and constant column added

The dashed vertical line in an augmented matrix separates coefficients from constants (sometimes omitted in other resources).
The augmented matrix contains all information needed to solve the system.

🔄 Converting between systems and matrices

System to matrix: Write coefficients as they appear in standard form, lined up by variable.
Matrix to system: Each row becomes an equation; columns correspond to variables in order.
Ensure equations are in standard form (variables on left, constant on right) before constructing the matrix.

⚙️ Elementary row operations

⚙️ The three allowed operations

Elementary row operations: Operations that can be performed to obtain equivalent linear systems.

Interchange: Any two rows may be swapped.
Scaling: Each element in a row can be multiplied by a nonzero constant.
Replacement: Any row can be replaced by the sum of that row and a multiple of another row.

These operations are consistent with the elimination method properties.
They produce equivalent systems (same solution set).
Why they work: They mirror the operations allowed when solving systems algebraically.

🎯 Goal of the operations

Transform the augmented matrix into upper triangular form.
Ideally, achieve row echelon form: upper triangular with each leading nonzero element equal to 1.
Once in this form, convert back to equations and use back substitution.

🔧 Gaussian elimination process

🔧 Step-by-step procedure

Gaussian elimination: Steps used to obtain an equivalent linear system in upper triangular form so that it can be solved using back substitution.

Step 1: Construct the augmented matrix from the system in standard form.

Step 2: Apply elementary row operations to obtain upper triangular form:

Eliminate elements below the first element of the first row (make them zero).
Then eliminate elements below the first nonzero element of the second row.
Continue until all elements below the diagonal are zero.

Step 3: Convert back to a linear system and solve using back substitution.

💡 Practical tips

It is more efficient if the leading element in each row is 1—consider interchanging rows or scaling to achieve this.
Work is typically done on scratch paper when replacing rows (multiply and add operations).
Many calculators and computer algebra systems can perform these operations automatically.

🔍 Example walkthrough (two variables)

Start with system: 9x - 6y = 0 and -x + 2y = 1.
Augmented matrix: first row [9, -6, 0], second row [-1, 2, 1].
Multiply second row by 9 and add to first row to eliminate the -1.
Result: first row unchanged, second row becomes [0, 12, 9].
Back substitute: 12y = 9 gives y = 3/4, then substitute into first equation to find x = 1/2.

🔀 Special cases: dependent and inconsistent systems

🔀 Recognizing dependent systems

A row of all zeros with a constant of zero indicates a dependent system.
The system has infinitely many solutions.
Solutions are expressed in terms of one or more free variables.
Example: If the third row becomes [0, 0, 0, 0], the system is dependent; express x and y in terms of z.

❌ Recognizing inconsistent systems

A row of all zeros with a nonzero constant indicates an inconsistent system.
The system has no solution.
This corresponds to an impossible equation like 0 = 5.

⚠️ Key distinction

Don't confuse: Both cases produce a row of zeros in the coefficient part, but:

Zero constant → dependent (infinitely many solutions)
Nonzero constant → inconsistent (no solution)

The constant column determines which case applies.

Historical note: The method is named after Carl Friedrich Gauss (1777–1855), though the excerpt notes that research into who first developed the process is encouraged.

Determinants and Cramer's Rule

Determinants and Cramer’s Rule

🧭 Overview

🧠 One-sentence thesis

Cramer's rule uses determinants of specially constructed matrices to solve linear systems efficiently, provided the coefficient matrix has a nonzero determinant.

📌 Key points (3–5)

What a determinant is: a real number calculated from a square matrix using specific formulas (subtraction of diagonal products for 2×2; expansion by minors for 3×3).
How Cramer's rule works: replace columns of the coefficient matrix with the constants column to form D_x, D_y (and D_z for three variables), then divide each by D.
When Cramer's rule applies: only when D ≠ 0; if D = 0, the system is either dependent (all numerator determinants also zero) or inconsistent (at least one numerator nonzero).
Common confusion: D = 0 does not automatically mean "no solution"—you must check whether D_x, D_y, D_z are also zero to distinguish dependent from inconsistent systems.
Why it matters: Cramer's rule provides a direct formula-based method for solving systems without row operations, especially useful for systems with three variables.

🔢 Understanding determinants

🔢 What a determinant is

Determinant: a real number associated with a square matrix.

A square matrix has the same number of rows and columns (e.g., 2×2 or 3×3).
The determinant is denoted |A| or det(A).
It is not the matrix itself; it is a single number computed from the matrix entries.

🧮 Calculating a 2×2 determinant

For a 2×2 matrix with entries a₁, a₂ (first column) and b₁, b₂ (second column):

The determinant is obtained by subtracting the products of the values on the diagonals: a₁b₂ − a₂b₁.

Example: For the matrix with first column (3, −5) and second column (2, −2), the determinant is 3(−2) − 2(−5) = −6 + 10 = 4.
The excerpt emphasizes: "subtract the product of the values on the diagonal."

🧩 Calculating a 3×3 determinant

For a 3×3 matrix, the determinant is calculated by expansion by minors about any row or column.

Minor: the determinant of the matrix that results after eliminating a row and column of a square matrix.

How expansion by minors works (using the first row):

Take each element in the first row.
Multiply it by the determinant of the 2×2 matrix formed by eliminating that element's row and column.
Alternate signs: + − + for the first row.
Formula: a₁(minor of a₁) − b₁(minor of b₁) + c₁(minor of c₁).

Sign array for any row or column:

The excerpt provides a sign pattern:

+ − +
− + −
+ − +

When expanding about the second row, start with − (opposite of the first element's sign in the array).
Example: Expanding about the second row of a matrix, the signs are − + −.

Why this matters:

You can expand about any row or column; the result is the same.
Choose a row or column with zeros to simplify calculations (the excerpt demonstrates this: "calculations are simplified if we expand about the third column because it contains two zeros").

Example: For a matrix with third column (0, 2, 0), expanding about that column means only the middle term contributes (the two zeros eliminate their terms).

🔧 Cramer's rule for two variables

🔧 The setup

Given a 2×2 linear system in standard form:

a₁x + b₁y = c₁
a₂x + b₂y = c₂

Form the augmented matrix and define three determinants:

Determinant	What it is	How to form it
D	Determinant of the coefficient matrix	Use the two columns of coefficients (a₁, a₂) and (b₁, b₂)
D_x	Determinant for solving x	Replace the first column (x-coefficients) with the constants (c₁, c₂)
D_y	Determinant for solving y	Replace the second column (y-coefficients) with the constants (c₁, c₂)

📐 Cramer's rule formula

Cramer's rule (two variables): (x, y) = (D_x / D, D_y / D), provided D ≠ 0.

Step-by-step process (from the excerpt):

Ensure the system is in standard form.
Construct the augmented matrix and form D, D_x, D_y.
Calculate the three determinants.
Apply Cramer's rule: x = D_x / D and y = D_y / D.
(Optional) Check the solution by substituting back.

Example: For the system 2x + y = 7, 3x − 2y = −7:

D = 2(−2) − 3(1) = −7
D_x = 7(−2) − (−7)(1) = −7
D_y = 2(−7) − 3(7) = −35
Solution: x = −7/(−7) = 1, y = −35/(−7) = 5.

⚠️ When D = 0

If D = 0, Cramer's rule formulas are undefined (division by zero).

The excerpt explains:

Condition	System type
D = 0 and both D_x = 0 and D_y = 0	Dependent (infinitely many solutions)
D = 0 and either D_x ≠ 0 or D_y ≠ 0	Inconsistent (no solution)

Don't confuse: D = 0 alone does not tell you whether the system is dependent or inconsistent—you must check the numerator determinants.
Example: If D = 0, D_x = 0, D_y = 0, the two equations represent the same line (dependent).
Example: If D = 0 but D_x ≠ 0, the system has no solution (inconsistent).

🔧 Cramer's rule for three variables

🔧 The setup for 3×3 systems

Given a 3×3 linear system:

a₁x + b₁y + c₁z = d₁
a₂x + b₂y + c₂z = d₂
a₃x + b₃y + c₃z = d₃

Form the augmented matrix and define four determinants:

Determinant	What it is
D	Determinant of the 3×3 coefficient matrix
D_x	Replace the first column (x-coefficients) with the constants (d₁, d₂, d₃)
D_y	Replace the second column (y-coefficients) with the constants
D_z	Replace the third column (z-coefficients) with the constants

📐 Cramer's rule formula (three variables)

Cramer's rule (three variables): (x, y, z) = (D_x / D, D_y / D, D_z / D), provided D ≠ 0.

Process:

Write the augmented matrix.
Calculate D using expansion by minors.
Calculate D_x, D_y, D_z (each by replacing one column with constants).
Apply the formulas.

Example: For a system with D = 11, D_x = −44, D_y = 0, D_z = −33:

x = −44/11 = −4
y = 0/11 = 0
z = −33/11 = −3
Solution: (−4, 0, −3).

⚠️ When D = 0 (three variables)

The same logic applies as for two variables:

Condition	System type
D = 0 and D_x = 0 and D_y = 0 and D_z = 0	Dependent
D = 0 and at least one of D_x, D_y, D_z is nonzero	Inconsistent

Example: If D = 0 but D_x = 96 ≠ 0, the system is inconsistent (no solution).

🎯 Key takeaways from the excerpt

🎯 When to use Cramer's rule

Efficient for small systems: The excerpt emphasizes that Cramer's rule "efficiently determines solutions to linear systems."
Direct formula method: No need for row operations (Gaussian elimination); you calculate determinants and apply formulas.
Limitation: Only works when D ≠ 0; otherwise, you must analyze D_x, D_y, D_z to classify the system.

🎯 Important reminders

The determinant is a real number, not a matrix.
For 2×2: subtract diagonal products.
For 3×3: expand by minors about any row or column, using the sign array.
Cramer's rule does not apply when D = 0; in that case, check whether the system is dependent or inconsistent by examining the numerator determinants.
Modern calculators and computer algebra systems can compute determinants; the excerpt encourages researching this topic further.

Solving Systems of Inequalities with Two Variables

🧭 Overview

🧠 One-sentence thesis

A system of inequalities with two variables is solved by finding the intersection of the solution regions of each inequality, which represents all ordered pairs that satisfy all inequalities simultaneously.

📌 Key points (3–5)

What a system of inequalities is: a set of two or more inequalities with the same variables that must be considered simultaneously.
How solutions work: each inequality has infinitely many ordered pair solutions forming a region; the system's solution is the intersection of these regions.
How to find solutions graphically: graph each inequality separately, then identify where the shaded regions overlap.
How to verify solutions: check that a candidate ordered pair satisfies all inequalities in the system.
Common confusion: the solution is not just one inequality's region—it must be in the intersection where all conditions hold at once.

📐 Understanding systems of inequalities

📐 What a system of inequalities means

A system of inequalities: a set of two or more inequalities with the same variables.

The inequalities define conditions that must be considered simultaneously.
Unlike a single inequality, you are looking for ordered pairs that satisfy all inequalities at the same time.
Example from the excerpt: the system consists of y > x − 2 and y ≤ 2x + 2 together.

🔢 Solution sets and regions

Each individual inequality contains infinitely many ordered pair solutions.
These solutions are defined by a region in a rectangular coordinate plane.
When considering two inequalities together, the intersection of these regions defines the set of simultaneous solutions.
Don't confuse: the solution to the system is not the union (all solutions to either inequality), but the intersection (solutions to both).

🎨 Graphing solution sets

🎨 The graphing process

The excerpt describes a step-by-step approach:

Graph each inequality separately on its own coordinate plane.
Graph both on the same set of axes to see where regions overlap.
Identify the intersection where both conditions are met.

🖍️ Visual representation

The intersection is typically shaded darker to distinguish it from the individual regions.
The final graph presents only the intersection as the solution set.
This visual method makes it easy to see which ordered pairs satisfy all inequalities.

✅ Verifying solutions

✅ How to check a solution

The excerpt shows that (3, 2) appears to be a solution because it lies in the intersection region.
To verify, substitute the ordered pair into each original inequality.
The point is a valid solution only if it satisfies both (or all) inequalities in the system.

🔍 Why verification matters

The graph provides a visual estimate, but algebraic verification confirms the solution.
Example: for the point (3, 2) in the system with y > x − 2 and y ≤ 2x + 2, you would check:
- Does 2 > 3 − 2? (checking the first inequality)
- Does 2 ≤ 2(3) + 2? (checking the second inequality)
Only if both are true is (3, 2) a solution to the system.

Algebraic Expressions and Formulas

Algebra of Functions

🧭 Overview

🧠 One-sentence thesis

Algebraic expressions combine variables and numbers through operations, and simplifying them requires applying the distributive property and combining like terms before evaluating them by substituting specific values.

📌 Key points (3–5)

What algebraic expressions are: combinations of variables (letters representing numbers) and numbers joined by mathematical operations.
How to simplify: apply the distributive property to remove parentheses, then combine like terms (terms with identical variable parts).
Key structure: terms are separated by addition; each term has a coefficient (numerical factor) and a variable part (all variable factors with exponents).
Common confusion: like terms must have exactly the same variable parts (same variables with same exponents)—only then can you add/subtract coefficients.
Evaluating expressions: substitute given values for variables (use parentheses first), then simplify using order of operations.

🧩 Building blocks of expressions

🧩 What makes up an expression

Algebraic expression: combinations of variables and numbers along with mathematical operations used to generalize specific arithmetic operations.

Variables are letters (commonly x and y, but any letter or Greek letters like α, β) that represent unknown numbers.
Example expressions: 2x + 3, x squared minus 9, 1 divided by x plus x divided by (x + 2).

🔢 Terms, coefficients, and variable parts

Terms: components of an algebraic expression separated by addition operators.

Coefficient: the numerical factor of a term.

Variable part of a term: all the variable factors with their exponents.

Factors are separated by multiplication operators within a term.
Example: in x squared y squared + 6xy − 3:
- First term: x squared y squared has coefficient 1 and variable part x squared y squared.
- Second term: 6xy has coefficient 6 and variable part xy.
- Third term: −3 is a constant term (no variable factor).
Don't confuse: a term like −b squared should be thought of as −1 times b squared (coefficient is −1, not missing).

📋 Identifying components

Term	Coefficient	Variable Part
10a squared	10	a squared
−5ab	−5	ab
−b squared	−1	b squared

Constant terms (numbers without variables) are also terms; their coefficient is the number itself.

🔄 The distributive property

🔄 What the distributive property says

Distributive property: given any real numbers a, b, and c, a times (b + c) equals ab + ac, or (b + c) times a equals ba + ca.

This property lets you multiply a factor outside parentheses by every term inside.
Why it matters: when you cannot simplify inside parentheses further, use the distributive property to remove parentheses.

🛠️ Applying the distributive property

Multiply the outside factor by each term inside the parentheses.
Example: 5 times (−2a + 5b) − 2c
- Distribute 5: 5 times (−2a) + 5 times 5b − 2c
- Result: −10a + 25b − 2c
Example: (3x − 4y + 1) times 3
- Distribute 3: 3x times 3 − 4y times 3 + 1 times 3
- Result: 9x − 12y + 3
Don't confuse: only multiply terms inside the parentheses by the outside factor; terms outside (like −2c above) are not affected.

🧲 Combining like terms

🧲 What are like terms

Like terms (similar terms): constant terms or terms whose variable parts have the same variables with the same exponents.

The variable parts must be exactly the same: same letters, same exponents.
Example: 5x and 7x are like terms (both have variable part x).
Example: 4x squared and 5x squared are like terms (both have variable part x squared).
Example: 12x squared y cubed and 3x squared y cubed are like terms.

➕ How to combine like terms

Add or subtract the coefficients; keep the variable part unchanged.
Example: 5x + 7x = (5 + 7)x = 12x
Example: 4x squared + 5x squared − 7x squared = (4 + 5 − 7)x squared = 2x squared
Example: 12x squared y cubed + 3x squared y cubed = 15x squared y cubed
Don't confuse: variable factors and exponents do not change when combining like terms.

🧹 Simplifying expressions

Simplifying the expression: combining like terms until the expression contains no more similar terms.

Identify all like terms, then add their coefficients.
Example: x squared − 10x + 8 + 5x squared − 6x − 1
- Like terms: x squared and 5x squared → 6x squared
- Like terms: −10x and −6x → −16x
- Like terms: 8 and −1 → 7
- Result: 6x squared − 16x + 7
Example with distributive property first: a squared b squared − ab − 2(2a squared b squared − 5ab + 1)
- Distribute −2: a squared b squared − ab − 4a squared b squared + 10ab − 2
- Combine like terms: (1 − 4)a squared b squared + (−1 + 10)ab − 2
- Result: −3a squared b squared + 9ab − 2

🔢 Evaluating algebraic expressions

🔢 What evaluating means

Evaluating: the process of performing the operations of an algebraic expression for given values of the variables.

Replace each variable with its assigned number, then compute using order of operations.
Best practice: first replace all variables with parentheses, then substitute the values.

🔄 Substitution steps

Substitute: the act of replacing a variable with an equivalent quantity.

Step 1: Write the expression with parentheses where each variable appears.
Step 2: Replace parentheses with the given values.
Step 3: Simplify using order of operations.

🧮 Single-variable examples

Example: evaluate 5x − 2 where x = two-thirds
- Replace: 5( ) − 2
- Substitute: 5(two-thirds) − 2
- Simplify: ten-thirds − six-thirds = four-thirds
Example: evaluate y squared − y − 6 where y = −4
- Replace: ( ) squared − ( ) − 6
- Substitute: (−4) squared − (−4) − 6
- Simplify: 16 + 4 − 6 = 14
Don't confuse: use parentheses to avoid sign errors, especially with negative values.

🔢 Multi-variable examples

Example: evaluate a cubed − 8b cubed where a = −1 and b = one-half
- Replace: ( ) cubed − 8( ) cubed
- Substitute: (−1) cubed − 8(one-half) cubed
- Simplify: −1 − 8(one-eighth) = −1 − 1 = −2
Order of operations is critical: exponents before multiplication, multiplication before addition/subtraction.

Evaluating Algebraic Expressions and Formulas

Factoring Polynomials

🧭 Overview

🧠 One-sentence thesis

Algebraic expressions generalize arithmetic operations by using variables, and evaluating them requires substituting given values and simplifying using the order of operations, which enables reusable formulas for real-world applications.

📌 Key points (3–5)

What evaluating means: replacing variables with given numbers and performing the operations using the order of operations.
Best practice to avoid errors: replace all variables with parentheses first, then substitute values.
Formulas as reusable models: algebraic expressions allow us to create formulas that describe common applications (volume, distance, interest, etc.).
Common confusion: improper fractions vs mixed numbers—solutions are expressed as reduced improper fractions unless the original problem uses mixed numbers or is a real-world application.
Why it matters: formulas model relationships between quantities and let us calculate unknown values efficiently.

🔢 What evaluating means

🔢 Definition and process

Evaluating: performing the operations of an algebraic expression for given values of the variables.

An algebraic expression is a generalization of arithmetic operations using variables.
When a problem assigns a specific value to a variable, you replace the letter with that number.
Then simplify using the order of operations.

🛡️ Best practice: parentheses first

Why: to avoid common errors.
How: first replace all variables with parentheses, then substitute the appropriate given value.
Example: To evaluate 5x − 2 where x = 2/3, write 5( ) − 2, then 5(2/3) − 2, then simplify to 10/3 − 2 = 4/3.

🔄 Substitution

Substitute: the act of replacing a variable with an equivalent quantity.

After placing parentheses, insert the given number in place of each variable.
Then perform arithmetic operations step by step.

🧮 Evaluating with multiple variables

🧮 Multiple variables

Algebraic expressions often involve more than one variable.
Example: Evaluate a³ − 8b³ where a = −1 and b = 1/2.
- Replace: ( )³ − 8( )³
- Substitute: (−1)³ − 8(1/2)³
- Simplify: −1 − 8(1/8) = −1 − 1 = −2.

🧩 Complex fractions

When the expression is a fraction, simplify the numerator and denominator separately.
Then multiply by the reciprocal of the denominator.
Example: Evaluate (x² − y²)/(2x − 1) where x = −3/2 and y = −3.
- Numerator: (−3/2)² − (−3)² = 9/4 − 9 = −27/4.
- Denominator: 2(−3/2) − 1 = −3 − 1 = −4.
- Result: (−27/4) ÷ (−4) = (−27/4)(−1/4) = 27/16.

🔢 Improper fractions vs mixed numbers

Common confusion: when to use mixed numbers vs improper fractions.
Rule from the excerpt: unless the original problem has mixed numbers or it is a real-world application, express solutions as reduced improper fractions.
Example: 27/16 can be written as 1 11/16, but leave it as 27/16 unless specified otherwise.

🧮 Square roots and radicals

When evaluating expressions with square roots, substitute values, simplify under the radical, then simplify the radical itself.
Example: Evaluate the square root of (b² − 4ac) where a = −1, b = −7, c = 1/4.
- Substitute: square root of [(−7)² − 4(−1)(1/4)]
- Simplify: square root of [49 + 1] = square root of 50
- Factor: square root of (25 · 2) = 5 times the square root of 2.

📐 Using formulas

📐 What formulas are

Formulas: reusable mathematical models using algebraic expressions to describe common applications.

The main difference between algebra and arithmetic is the organized use of variables.
Variables and constants describe relationships between quantities.
Example: The volume of a right circular cone is V = (1/3)πr²h, where r is radius and h is height.

📏 Geometry formulas

Perimeter (P): measured in linear units.
Area (A): measured in square units.
Surface area (SA): measured in square units.
Volume (V): measured in cubic units.
The excerpt provides lists of formulas for common plane figures and three-dimensional figures.

🎈 Example: Volume of a sphere

Formula: V = (4/3)πr³.
The formula gives volume in terms of radius r.
Example: A spherical balloon has diameter 10 inches. Find the volume.
- Radius r = 10/2 = 5 inches.
- Substitute: V = (4/3)π(5)³ = (4/3)π(125) = (500π)/3 cubic inches.
- Approximate using π ≈ 3.14: V ≈ 523.60 cubic inches.

🚗 Real-world applications

🚗 Uniform motion

Uniform motion: the distance D after traveling at an average rate r for some time t can be calculated using the formula D = rt.

Read as "distance equals rate times time."
Example: A road trip took 2 1/2 hours at an average speed of 66 miles per hour. How far?
- Substitute: D = 66 · (5/2) = 33 · 5 = 165 miles.

💰 Simple interest

Simple interest: modeled by the formula I = prt, where p represents the principal amount invested at an annual interest rate r for t years.

p: principal (initial amount).
r: annual interest rate (as a decimal).
t: time in years.
Example: Calculate simple interest on a 2-year investment of $1,250 at 3 3/4% annual rate.
- Convert rate: 3 3/4% = 3.75% = 0.0375.
- Substitute: I = 1,250 · 0.0375 · 2 = 93.75.
- Answer: $93.75 in interest earned.

🔑 Key takeaways from the excerpt

🔑 Core ideas

Algebraic expressions: generalizations of common arithmetic operations formed by combining numbers, variables, and mathematical operations.
Distributive property: a(b + c) = ab + ac, used when multiplying grouped expressions; applying it removes parentheses.
Combining like terms: terms whose variable parts have the same variables with the same exponents; add or subtract coefficients; variable factors and exponents do not change.
Evaluating best practice: replace all variables with parentheses, then substitute values to avoid errors.
Formulas: algebraic expressions allow creation of useful, reusable formulas that model common applications.

Algebraic Expressions and Formulas

Factoring Trinomials

🧭 Overview

🧠 One-sentence thesis

Algebraic expressions generalize arithmetic operations by combining numbers, variables, and operations, and they enable the creation of reusable formulas that model common applications.

📌 Key points (3–5)

What algebraic expressions are: generalizations of arithmetic operations formed by combining numbers, variables, and mathematical operations.
The distributive property: used to multiply grouped expressions and remove parentheses by applying a(b + c) = ab + ac.
Combining like terms: add or subtract coefficients of terms with identical variable parts (same variables with same exponents); the variable factors and exponents do not change.
Common confusion: when evaluating expressions, forgetting to use parentheses when substituting values can lead to errors—always replace variables with parentheses first.
Why formulas matter: algebraic expressions allow us to create useful, reusable formulas that model common applications like interest calculations and geometric measurements.

🔧 Building and simplifying expressions

🔧 What algebraic expressions are

Algebraic expressions: generalizations of common arithmetic operations formed by combining numbers, variables, and mathematical operations.

They extend basic arithmetic to work with unknown or variable quantities.
Example: Instead of calculating 5 times 3 minus 5, you can write 5x - 5 to represent 5 times any number x minus 5.

🔀 The distributive property

The distributive property: a(b + c) = ab + ac, used when multiplying grouped algebraic expressions.

Purpose: Applying the distributive property allows us to remove parentheses.
How it works: Multiply the term outside the parentheses by each term inside.
Example: 5(3x - 5) becomes 5 times 3x plus 5 times negative 5, which equals 15x - 25.
Example: negative 2(2x squared - 5x + 1) becomes negative 4x squared + 10x - 2.

➕ Combining like terms

Like terms: terms whose variable parts have the same variables with the same exponents.

How to combine: Add or subtract the coefficients to obtain the coefficient of a single term with the same variable part.
Key rule: The variable factors and their exponents do not change.
Example: 18x - 5x + 3x combines to 16x (only the coefficients 18, -5, and 3 are added).
Example: 2x squared - 3x + 2 + 5x squared - 6x + 1 becomes 7x squared - 9x + 3 (combine 2x squared with 5x squared, -3x with -6x, and 2 with 1).
Don't confuse: You cannot combine 3x squared with 5x because the exponents differ; you cannot combine 5xy with 2xy squared because the variable parts differ.

🧮 Evaluating expressions

🧮 Substitution best practice

Best practice: Replace all variables with parentheses and then substitute the appropriate values.

Why: To avoid common errors when evaluating.
How: Before plugging in numbers, put parentheses where each variable appears.
Example: To evaluate negative 2x + 3 where x = negative 2, write negative 2(x) + 3, then substitute to get negative 2(negative 2) + 3 = 4 + 3 = 7.
Common error: Writing negative 2 times negative 2 without parentheses can lead to sign mistakes.

🔢 Working with specific values

The excerpt shows evaluating expressions by substituting given values:

For x squared - x + 5 where x = negative 5: substitute to get (negative 5) squared - (negative 5) + 5 = 25 + 5 + 5 = 35.
For expressions with multiple variables like a squared - 5b squared where a = negative 2 and b = negative 1: substitute both values and compute (negative 2) squared - 5(negative 1) squared = 4 - 5(1) = 4 - 5 = negative 1.

📐 Using formulas

📐 Simple interest formula

Simple interest formula: I = prt, where p represents the principal amount invested at an annual interest rate r for t years.

Components:
- I = interest earned
- p = principal (initial amount)
- r = annual interest rate (as a decimal)
- t = time in years
Example: Calculate simple interest on a 2-year investment of $1,250 at an annual interest rate of 3 and 3/4 percent.
- Convert 3 and 3/4 percent to decimal: 3.75 percent = 0.0375
- Substitute: I = (1,250)(0.0375)(2) = 93.75
- The simple interest earned is $93.75.

📏 Geometric formulas

The excerpt mentions formulas for:

Shape	Formulas mentioned
Rectangle	Perimeter and area (with dimensions given)
Sphere	Surface area and volume (with radius given)

These formulas are reusable: once you know the formula, you can apply it to any rectangle or sphere by substituting the specific measurements.

🌡️ Temperature conversion

Temperature conversion formula: C = 5/9(F - 32), where F represents degrees Fahrenheit and C represents degrees Celsius.

How to use: Substitute the Fahrenheit temperature for F and calculate C.
Example: Convert 95°F to Celsius: C = 5/9(95 - 32) = 5/9(63) = 35°C.
Example: Convert 32°F to Celsius: C = 5/9(32 - 32) = 5/9(0) = 0°C.

Solve Polynomial Equations by Factoring

🧭 Overview

🧠 One-sentence thesis

This excerpt provides practice exercises for evaluating algebraic expressions and applying formulas, but does not contain substantive instructional content on solving polynomial equations by factoring.

📌 Key points (3–5)

The excerpt consists primarily of numbered practice problems for evaluating expressions at given variable values.
It includes exercises on using formulas for temperature conversion, geometry (perimeter, area, volume), distance-speed-time, and simple interest.
The material covers algebraic expression simplification and evaluation, not the factoring methods suggested by the title.
Answer keys are provided at the end for self-checking.
Discussion board prompts encourage exploration of mathematical models and the distributive property.

📝 Content mismatch

📝 What the excerpt actually contains

The excerpt does not teach methods for solving polynomial equations by factoring. Instead, it presents:

Practice problems numbered 60–110 for evaluating algebraic expressions when variables are assigned specific numerical values.
Formula application exercises (temperature conversion, geometric measurements, motion problems, simple interest).
A brief transition at the very end mentioning "Review of the Rules of Exponents" as the start of section 1.5.

⚠️ Missing instructional content

No explanation of factoring techniques (greatest common factor, difference of squares, trinomial factoring, grouping).
No worked examples showing how to set a polynomial equal to zero and solve by factoring.
No discussion of the zero-product property or how factoring leads to solutions.

🧮 Exercise categories present

🧮 Expression evaluation

The bulk of problems (60–82) ask students to substitute given values into algebraic expressions and compute the result.

Example format: Evaluate "9x² + x - 2" where x = -2/3.
Some expressions involve two variables (a and b, or x and y).
Problems 75–82 specifically evaluate the discriminant formula: the square root of (b² - 4ac).

📐 Formula applications

Problems 83–104 apply standard formulas:

Formula type	Examples given
Temperature	Celsius = 5/9(F - 32)
Geometry	Perimeter, area, surface area, volume for rectangles, spheres, cylinders, cones
Motion	Distance = speed × time
Finance	Simple interest
Physics	Free fall time, electrical current

💬 Discussion prompts

Problems 105–110 are open-ended questions asking students to research mathematical models, the history of variables, the Greek alphabet, and explain properties like distribution.

📋 Answer key structure

📋 Solutions provided

The excerpt includes answers for odd-numbered problems, allowing self-checking:

Simplified algebraic expressions (problems 1–53).
Numerical evaluations (problems 55–73).
Exact and approximate answers for formulas (e.g., "2√2 ≈ 1.41 seconds").
"Answer may vary" for discussion questions.

Rational Functions: Multiplication and Division

🧭 Overview

🧠 One-sentence thesis

This excerpt does not contain substantive content on rational functions, multiplication, or division; instead, it consists entirely of practice problems on algebraic expressions, formulas, and exponents, along with their answers.

📌 Key points (3–5)

The excerpt is a collection of exercise problems (numbered 60–110) from a textbook chapter on algebraic expressions and formulas.
Problems cover evaluating expressions at given variable values, using formulas (temperature conversion, geometry, distance/speed/time, simple interest), and discussion prompts.
The excerpt includes answer keys for the problems but no explanatory text or worked examples.
The final visible section begins introducing exponent rules but is cut off mid-sentence.
Common confusion: The title "Rational Functions: Multiplication and Division" does not match the content, which focuses on algebraic expression evaluation and formula application.

📝 Content summary

📝 What the excerpt contains

The excerpt is structured as:

Part B: Evaluation problems (numbered 60–82) asking students to substitute specific values into algebraic expressions and compute results.
Part C: Formula application problems (numbered 83–104) covering temperature conversion, geometry (perimeter, area, volume), distance-speed-time, and simple interest.
Part D: Discussion board prompts (numbered 105–110) asking students to research topics like mathematical models, the history of variables, and the Greek alphabet.
Answers section: Numerical or algebraic answers to the problems.
Section 1.5 preview: A brief introduction to exponent rules that is incomplete.

🔢 Types of problems included

The excerpt does not teach concepts but instead provides practice exercises:

Substituting values into polynomial and rational expressions.
Evaluating the quadratic discriminant formula (b squared minus 4ac, under a square root).
Converting Fahrenheit to Celsius using the given formula.
Calculating geometric properties (perimeter, area, surface area, volume) for rectangles, spheres, cylinders, and cones.
Solving distance-speed-time problems.
Computing simple interest.

⚠️ What is missing

No instructional text explaining how to multiply or divide rational functions.
No definitions, theorems, or worked examples related to rational functions.
No explanation of the algebraic techniques needed to solve the problems.
The exponent rules section (1.5) is cut off and does not provide complete information.

🧮 Problem categories

🧮 Expression evaluation

Problems 60–74 ask students to substitute given values into algebraic expressions and simplify. Examples include:

Expressions with negative exponents and fractional bases.
Polynomial expressions in one or two variables.
Difference-of-squares patterns.

Example: Problem 61 asks to evaluate (3y − 2)(y + 5) where y = 2/3.

📐 Formula application

Problems 75–104 require using standard formulas:

Formula type	What students must do
Quadratic discriminant	Evaluate the square root of (b squared minus 4ac) for given a, b, c values
Temperature conversion	Convert Fahrenheit to Celsius using C = 5/9(F − 32)
Geometry	Calculate perimeter, area, surface area, or volume for given shapes and dimensions
Distance-speed-time	Use distance = speed × time to find missing quantities
Simple interest	Calculate interest earned using principal, rate, and time

💬 Discussion prompts

Problems 105–110 are open-ended questions asking students to:

Find and share mathematical models or resources.
Explain order-of-operations rules (e.g., why we don't subtract 5 and 3 first in 5 − 3(9x − 1)).
Discuss whether the distributive property extends to more than two terms.

🔍 Relationship to title

🔍 Mismatch between title and content

The title "Rational Functions: Multiplication and Division" suggests the excerpt should explain:

What rational functions are (ratios of polynomials).
How to multiply rational functions (multiply numerators and denominators, then simplify).
How to divide rational functions (multiply by the reciprocal).

However, the excerpt contains none of this material. It is instead a problem set from an earlier chapter on algebraic expressions and formulas, with no mention of rational functions.

🔍 Incomplete exponent section

The excerpt ends mid-sentence in Section 1.5 ("Rules of Exponents and Scientific Notation"), which begins to explain:

"The positive integer exponent n indicates the number of times the base x is repeated as a factor."

The text starts to illustrate that multiplying x to the fourth power by x to the sixth power can be simplified by adding exponents, but the explanation is cut off. This section is also unrelated to the title topic.

Rational Functions: Addition and Subtraction

🧭 Overview

🧠 One-sentence thesis

Adding and subtracting rational functions follows the same principles as fraction arithmetic—requiring common denominators—and complex rational expressions can be simplified by either division or LCD multiplication methods.

📌 Key points (3–5)

Same-denominator case: When rational expressions share a denominator, add or subtract numerators directly and write over the common denominator.
Different-denominator case: Find the least common denominator (LCD), convert to equivalent fractions, then combine.
Domain restrictions: The domain of a sum or difference includes all restrictions from each individual rational expression.
Common confusion: Don't forget to distribute negatives when subtracting numerators; the entire second numerator must be subtracted.
Complex rational expressions: Two methods exist—simplify numerator/denominator separately then divide, or multiply everything by the LCD to clear fractions.

➕ Adding and subtracting with the same denominator

➕ The basic rule

When rational expressions have the same denominator Q (where Q ≠ 0), combine them as: P/Q ± R/Q = (P ± R)/Q

Only the numerators are combined; the denominator stays the same.
This mirrors fraction addition: 2/5 + 3/5 = 5/5.

⚠️ Watch for subtraction

When subtracting, the entire second numerator must be subtracted, not just the first term.

Example: (4x)/(x² - 64) - (3x + 8)/(x² - 64)

Write as: [4x - (3x + 8)]/(x² - 64)
Distribute the negative: (4x - 3x - 8)/(x² - 64) = (x - 8)/(x² - 64)
Factor and simplify: (x - 8)/[(x + 8)(x - 8)] = 1/(x + 8), where x ≠ ±8

Don't confuse: Writing 4x - 3x + 8 (forgetting to distribute the negative) is a common error.

🔄 Adding and subtracting with different denominators

🔍 Finding the LCD

When denominators differ, you must find equivalent expressions with a common denominator.

If denominators are relatively prime (share no common factors), the LCD is their product.
More typically, factor all denominators completely; the LCD is the product of all factors with their highest powers.

🔧 The conversion process

For P/Q ± R/S (where Q ≠ 0 and S ≠ 0):

General formula: P/Q ± R/S = (PS ± QR)/(QS)
Multiply each fraction by an appropriate form of 1 to obtain the LCD.

Example: Given f(x) = 5x/(3x + 1) and g(x) = 2/(x + 1), find f + g.

LCD = (3x + 1)(x + 1)
Convert: [5x(x + 1)]/[(3x + 1)(x + 1)] + [2(3x + 1)]/[(x + 1)(3x + 1)]
Combine numerators: [5x² + 5x + 6x + 2]/[(3x + 1)(x + 1)] = (5x² + 11x + 2)/[(3x + 1)(x + 1)]
Factor if possible: (5x + 1)(x + 2)/[(3x + 1)(x + 1)]

📋 Domain restrictions

The domain of f + g or f - g consists of all real numbers except values that make any denominator zero.

In the example above: x ≠ -1 and x ≠ -1/3.

🔀 Opposite binomial property

When denominators differ only by sign, use: 6 - x = -(x - 6)

This allows you to rewrite fractions with a common denominator more easily.

🎯 Complex rational expressions

🎯 What they are

A complex rational expression: a rational expression containing one or more rational expressions in its numerator or denominator (or both).

Example: [4 - 12/x + 9/x²] / [2 - 5/x + 3/x²]

The goal is to simplify to a single fraction with polynomial numerator and denominator.

🛠️ Method 1: Simplify using division

Step 1: Simplify numerator and denominator separately into single fractions.

Find common denominators within the numerator.
Find common denominators within the denominator.

Step 2: Multiply the numerator by the reciprocal of the denominator.

Step 3: Factor all numerators and denominators completely.

Step 4: Cancel common factors.

Example: [4 - 12/x + 9/x²] / [2 - 5/x + 3/x²]

Rewrite with common denominator x²: [(4x² - 12x + 9)/x²] / [(2x² - 5x + 3)/x²]
Multiply by reciprocal: [(4x² - 12x + 9)/x²] · [x²/(2x² - 5x + 3)]
Factor: [(2x - 3)(2x - 3)/x²] · [x²/((2x - 3)(x - 1))]
Cancel: (2x - 3)/(x - 1)

🧮 Method 2: Simplify using the LCD

Step 1: Find the LCD of all fractions appearing anywhere in the complex expression.

Step 2: Multiply both numerator and denominator by this LCD.

This clears all fractions at once.

Step 3: Factor the resulting polynomials.

Step 4: Cancel common factors.

Example: Same problem as above—LCD of all fractions is x².

Multiply top and bottom by x²: [(4 - 12/x + 9/x²) · x²] / [(2 - 5/x + 3/x²) · x²]
Distribute: (4x² - 12x + 9)/(2x² - 5x + 3)
Factor and cancel as before.

Don't confuse: Multiplying numerator and denominator by the same nonzero expression equals multiplying by 1, so it doesn't change the value—only the form.

⚡ Negative exponents

When expressions use negative exponents, first rewrite using positive exponents (x⁻ⁿ = 1/xⁿ), then apply either method.

Example: (2y⁻¹ - x⁻¹)/(x⁻² - 4y⁻²)

Rewrite: (2/y - 1/x)/(1/x² - 4/y²)
Then simplify using either method.

📊 Key takeaways summary

Concept	Key principle
Same denominator	Combine numerators directly; keep denominator
Different denominators	Find LCD, convert, then combine
Domain	Union of all individual restrictions
Complex expressions—Method 1	Simplify top/bottom separately, then divide
Complex expressions—Method 2	Multiply by LCD to clear all fractions

Both methods for complex expressions yield the same result; choose based on which seems more efficient for the given problem.

Solving Rational Equations

🧭 Overview

🧠 One-sentence thesis

Rational equations are solved by multiplying both sides by the least common denominator to clear fractions, but solutions must always be checked against domain restrictions because multiplying by variable expressions can introduce extraneous solutions.

📌 Key points (3–5)

What a rational equation is: an equation containing at least one rational expression (fraction with variables in the denominator).
Core solving method: multiply both sides by the LCD to eliminate fractions, then solve the resulting equation (which may be linear or quadratic).
Critical checking step: always verify solutions against restrictions (values that make denominators zero); solutions that violate restrictions are extraneous and must be discarded.
Common confusion: don't confuse solving equations with simplifying expressions—clearing fractions by multiplying by the LCD only works for equations, not expressions.
Special cases: some rational equations have no solution (all candidates are extraneous, or the equation is a contradiction).

🔧 The clearing-fractions method

🔧 Step-by-step process

The excerpt outlines a systematic five-step approach:

Factor all denominators and determine the LCD
Identify restrictions (values that make any denominator zero)
Multiply both sides by the LCD, distributing carefully
Solve the resulting equation (may be linear, quadratic, or higher-degree)
Check for extraneous solutions by substituting back into the original equation

🎯 Why this works

Multiplying by the LCD transforms the rational equation into a polynomial equation (no fractions).
Example: For 1/x + 2/x² = (x+9)/(2x²), the LCD is 2x². Multiplying through eliminates all denominators, leaving 2x + 4 = x + 9, which simplifies to x = 5.

⚠️ The distribution requirement

When multiplying by the LCD, you must distribute to every term on both sides.
Don't confuse: multiplying one side only, or forgetting to distribute to all terms, produces incorrect results.

🚫 Restrictions and extraneous solutions

🚫 What restrictions are

Restrictions: values of the variable that make any denominator equal to zero.

These values are not in the domain of the original equation.
Must be identified before solving (from the factored denominators).
Example: For an equation with denominator (x - 4)(x - 2), the restrictions are x ≠ 4 and x ≠ 2.

👻 Extraneous solutions explained

Extraneous solutions: solutions that do not solve the original equation.

Why they occur: Multiplying both sides by a variable expression can introduce solutions that weren't valid for the original equation.
How to identify: Any solution that equals a restriction is automatically extraneous.
Example: If solving produces x = 5 but 5 is a restriction, then 5 is extraneous and must be rejected.

🔍 Two ways to have no solution

The excerpt demonstrates two scenarios:

Scenario	What happens	Example outcome
Contradiction	After clearing fractions, you get a false statement like `18 = 16`	No solution, Ø
All extraneous	Solving produces candidates, but all are restrictions	No solution, Ø

⚖️ Equations vs expressions (critical distinction)

⚖️ The fundamental difference

Expression: to be simplified (e.g., 1/x + x/(2x+1))
Equation: to be solved (e.g., 1/x + x/(2x+1) = 0)

🚨 Why clearing fractions doesn't work for expressions

The excerpt explicitly warns against this common error:

Incorrect: Multiplying an expression by the LCD changes it into a non-equivalent expression.
Correct: Only equations can have both sides multiplied by the LCD.
Example: 1/x + x/(2x+1) multiplied by x(2x+1) gives 2x+1 + x², which is not equivalent to the original expression.
Don't confuse: The same operation on an equation 1/x + x/(2x+1) = 0 is valid because you're maintaining equality.

🔄 Special techniques and cases

🔄 Negative exponents

Rational equations may use negative exponents: x⁻¹ means 1/x, x⁻² means 1/x².
Rewrite using positive exponents first, then apply the standard method.
Example: 6 + x⁻¹ = x⁻² becomes 6 + 1/x = 1/x².

✖️ Cross multiplication (proportions)

Proportion: a statement of equality of two ratios, a/b = c/d.

Cross multiplication rule: If a/b = c/d, then ad = bc.
This is a shortcut when each side of the equation is a single fraction.
Example: 5n - 1)/5 = 3n/2 cross-multiplies to 2(5n - 1) = 5(3n).
Don't confuse: Cross multiplication only works when you have exactly one fraction on each side; otherwise, use the LCD method.

🧮 Quadratic results

After clearing fractions, you may get a quadratic equation.
Factor and set each factor equal to zero.
Check both solutions against restrictions.
Example: An equation might yield x = 0 and x = -4; both must be checked to ensure neither is a restriction.

📐 Literal equations and applications

📐 Solving for a variable in formulas

Literal equations (formulas) with rational expressions use the same technique.
Multiply by the LCD to clear fractions, then isolate the desired variable.
Example: For 1/R = 1/R₁ + 1/R₂, multiply by RR₁R₂ to get R₁R₂ = RR₂ + RR₁, then factor and solve for R = R₁R₂/(R₁ + R₂).

🔢 Reciprocal applications

Reciprocal of a nonzero number n: the value 1/n such that n · (1/n) = 1.

Word problems involving "reciprocal" lead to rational equations.
Example setup: "The reciprocal of the smaller integer is subtracted from twice the reciprocal of the larger" translates to 2/n - 1/(n-3).
These problems may have multiple valid solutions (e.g., two different pairs of integers that satisfy the conditions).

✅ Verification in applications

Always check that solutions make sense in context (e.g., positive integers when specified).
Perform the operations described in the problem to verify the result.
Example: If the problem states a result should be 1/20, substitute your answer and confirm you get 1/20.

Applications and Variation

🧭 Overview

🧠 One-sentence thesis

Rational equations model real-world problems involving uniform motion, work rates, and variation relationships, where the algebraic setup often requires finding a constant of proportionality and solving for unknown quantities.

📌 Key points (3–5)

Uniform motion problems use the formula D = rt, rearranged as t = D/r when time is unknown, leading to rational equations.
Work-rate problems multiply individual work rates by time to find the portion of a task completed; the sum of portions equals the total work.
Direct variation describes y = kx (y varies directly as x), while inverse variation describes y = k/x (y varies inversely as x).
Common confusion: Direct vs inverse variation—direct means both quantities increase together; inverse means one increases as the other decreases.
Joint variation extends these concepts to multiple variables: y varies jointly as x and z means y = kxz.

🚗 Uniform motion applications

🚗 The basic formula and setup

Uniform motion (distance) problems: described by D = rt, where distance D is the product of average rate r and time t.

When time is the unknown, rearrange to t = D/r.
This rearrangement creates rational expressions, leading to rational equations.
Organization strategy: Use a chart with columns for distance, rate, and time.

🧮 Solving distance problems

Step-by-step approach:

Identify unknowns (let x represent one rate; express other rates in terms of x).
Fill in the chart using t = D/r for the time column.
Set up an equation based on the relationship between times (e.g., total time, equal times).
Multiply both sides by the LCD to clear denominators.
Solve the resulting polynomial equation.
Check that solutions make sense in context (discard negative speeds, etc.).

Example scenario: A person travels one distance by bus at speed x, then another distance by train at speed x + 18. If the total trip takes 2 hours, the setup is: (15/x) + (72/(x+18)) = 2.

🌊 Current and wind problems

With the current/wind: effective speed = vehicle speed + current speed.
Against the current/wind: effective speed = vehicle speed − current speed.
Common setup: equal time downstream and upstream, so set the two time expressions equal.

Example: A boat averages 12 mph in still water. Traveling 29 miles downstream and 19 miles upstream takes the same time. Setup: 29/(12+c) = 19/(12−c), where c is the current speed.

Don't confuse: The direction matters—add current speed when going with it, subtract when going against it.

🔧 Work-rate applications

🔧 Work rate fundamentals

Work rate: the rate at which a task can be performed, expressed as (1 task)/(time to complete).

If someone completes a task in 6 hours, their work rate is 1/6 of the task per hour.
Amount completed = work rate × time worked.
Example: Working for 3 hours at rate 1/6 completes (1/6)×3 = 1/2 of the task.

🤝 Combined work formula

Work-rate formula: (1/t₁)·t + (1/t₂)·t = amount of task completed together

Here 1/t₁ and 1/t₂ are individual work rates.
t is the time working together.
For one complete task: (1/t₁)·t + (1/t₂)·t = 1.

Equivalent forms:

t/t₁ + t/t₂ = 1
1/t₁ + 1/t₂ = 1/t

🛠️ Solving work problems

Setup strategy:

Let x = time for one person to complete the task alone.
Express the other person's time in terms of x (e.g., x − 2 if they're 2 hours faster).
Identify how long each person actually worked.
Use: (work rate₁)(time₁) + (work rate₂)(time₂) = total tasks completed.
Multiply by the LCD and solve.

Example: Joe paints a room in 2 hours less than Mark. Working together for 12 hours, they paint 5 rooms. If Mark takes x hours per room, Joe takes x − 2 hours. Setup: 12/(x−2) + 12/x = 5.

Don't confuse: Work rate (1/time) vs. time itself—the rate is the reciprocal of the time.

📊 Direct and inverse variation

📊 Direct variation

Direct variation: y = kx, where y varies directly as x and k is the constant of variation (or constant of proportionality).

Key phrases:

"y varies directly as x"
"y is directly proportional to x"
"y is proportional to x"

Characteristics:

As x increases, y increases proportionally.
The ratio y/x is constant (equals k).
Can extend to powers: y varies directly as x² means y = kx².

Example: Weight on Earth varies directly with weight on the Moon. If 180 lbs on Earth = 30 lbs on Moon, then k = 180/30 = 6, so y = 6x.

🔄 Inverse variation

Inverse variation: y = k/x, where y varies inversely as x and k is the constant of variation.

Key phrases:

"y varies inversely as x"
"y is inversely proportional to x"

Characteristics:

As x increases, y decreases.
The product xy is constant (equals k).
Can extend to powers: y varies inversely as x² means y = k/x².

Example: Light intensity I = k/d². At distance d = 1 foot, intensity = 525 foot-candles, so k = 525. Formula: I = 525/d².

Don't confuse direct vs inverse:

Direct: both increase/decrease together (y = kx).
Inverse: one increases while the other decreases (y = k/x).

🔗 Joint variation

Joint variation: y = kxz, where y varies jointly as x and z.

Key phrases:

"y varies jointly as x and z"
"y is jointly proportional to x and z"

Characteristics:

y is directly proportional to the product of two (or more) variables.
Example: Area of an ellipse varies jointly as a and b, so A = kab. Given A = 300π when a = 10 and b = 30, we find k = π, so A = πab.

🔍 Solving variation problems

🔍 Three-step solution process

Step 1: Set up the equation

Translate key words into an equation with constant k.
"varies directly" → y = kx
"varies inversely" → y = k/x
"varies jointly" → y = kxz

Step 2: Find the constant k

Use given values to solve for k.
Substitute known values of all variables into the equation.
Solve the resulting equation for k.

Step 3: Write the formula and solve

Substitute k back into the equation.
Use the formula to answer the question with new values.

🧪 Combined variation

Problems can combine direct and inverse variation:

"y varies directly as x and inversely as z" means y = kx/z.
"y varies directly as the square of x and inversely as z" means y = kx²/z.

Example: Given y varies directly as x² and inversely with z, where y = 2 when x = 3 and z = 27:

Setup: y = kx²/z
Find k: 2 = k(9)/27, so k = 6
Formula: y = 6x²/z
When x = 2 and z = 16: y = 6(4)/16 = 3/2

🎯 Key takeaways

🎯 Problem-solving strategies

For distance problems:

Use t = D/r to avoid extra variables.
Organize data in a chart before writing equations.
Remember to check that speeds are positive.

For work-rate problems:

Work rate × time = portion completed.
Sum of portions = total work done.
Individual rate = 1/(time to complete alone).

For variation problems:

Identify the type of variation from key words.
Always find k first using given information.
Write the complete formula before solving for unknowns.

⚠️ Common pitfalls

Distance problems: Forgetting to add/subtract current or wind speed correctly.
Work problems: Confusing work rate (1/t) with time (t).
Variation: Mixing up direct (y = kx) and inverse (y = k/x) relationships.
All types: Not checking restrictions (denominators ≠ 0, speeds > 0, times > 0).

Roots and Radicals

🧭 Overview

🧠 One-sentence thesis

Roots (square, cube, and nth roots) are operations that "undo" exponentiation, and simplifying radicals requires recognizing perfect powers and applying product/quotient rules while carefully handling even versus odd indices.

📌 Key points (3–5)

Square roots vs cube roots: square roots of positive numbers have two values (positive and negative), but the radical sign denotes only the principal (nonnegative) root; cube roots have only one real value and can be negative.
Even vs odd indices: when the index n is even, the nth root requires absolute value to ensure a nonnegative result and cannot be real if the radicand is negative; when n is odd, the nth root has the same sign as the radicand and no absolute value is needed.
Domain differences: square root functions require nonnegative radicands (domain restricted), while cube root functions accept all real numbers (domain unrestricted).
Common confusion: distinguishing when to use absolute value—use |a| when the index is even, but not when the index is odd.
Simplification strategy: factor the radicand to find the largest perfect power matching the index, then apply the product or quotient rule for radicals.

🔢 Square roots and their properties

🔢 Definition and principal root

Square root of a number: a number that when multiplied by itself yields the original number.

Every positive real number has two square roots: one positive, one negative.
Example: both 5 and −5 are square roots of 25 because 5² = 25 and (−5)² = 25.
The radical sign √ denotes the principal (nonnegative) square root only.
Zero has only one square root: √0 = 0.

🔢 Why absolute value matters

When simplifying √(a²), the result is |a|, not just a, because a might be negative.
Example: √((x − 2)²) = |x − 2|.
If x = 1, then √((1 − 2)²) = √(1) = 1, which equals |1 − 2| = |−1| = 1.
Don't confuse: √(a²) ≠ a in general; it equals |a| to ensure the principal root is nonnegative.

🚫 Square roots of negative numbers

The square root of a negative number is not a real number.
Example: √(−25) is undefined in the real number system because no real number squared gives −25.
This restriction affects the domain of square root functions.

📊 Square root and cube root functions

📊 Square root function domain and range

The square root function f(x) = √x is defined only when x ≥ 0.
Domain: [0, ∞) (nonnegative real numbers only).
Range: [0, ∞) (nonnegative outputs only).
To find the domain of f(x) = √(2x + 3), solve 2x + 3 ≥ 0, giving x ≥ −3/2, so domain is [−3/2, ∞).

📊 Cube root function domain and range

Cube root of a number: a number that when multiplied by itself three times yields the original number.

Notation: ∛a or the radical with index 3.
Any real number has exactly one real cube root (can be positive, negative, or zero).
Example: ∛64 = 4 because 4³ = 64; ∛(−64) = −4 because (−4)³ = −64.
Domain: (−∞, ∞) (all real numbers).
Range: (−∞, ∞) (all real numbers).
No absolute value needed: ∛(a³) = a for any real a.

📊 Graphical behavior

The square root function starts at (0, 0) and increases, staying in the first quadrant.
The cube root function passes through (0, 0) and extends into all quadrants, accepting negative inputs and producing negative outputs.

🔄 nth roots and index rules

🔄 General nth root definition

nth root: a number that when raised to the nth power (n ≥ 2) yields the original number.

Notation: ⁿ√(aⁿ) where n is the index and aⁿ is the radicand.
When n = 2, the index is usually omitted: √a.
Example: ⁴√81 = 3 because 3⁴ = 81.

🔄 Even index (n = 2, 4, 6, ...)

The principal nth root is nonnegative: ⁿ√(aⁿ) = |a| when n is even.
Example: ⁴√((−3)⁴) = ⁴√81 = |−3| = 3.
Negative radicands are not real: ⁴√(−81) is not a real number because no real number raised to an even power gives a negative result.
Don't confuse: −⁴√81 = −3 (negative sign outside) vs ⁴√(−81) (not real).

🔄 Odd index (n = 3, 5, 7, ...)

The nth root has the same sign as the radicand: ⁿ√(aⁿ) = a when n is odd (no absolute value).
Example: ⁵√((−10)⁵) = −10.
Negative radicands are real: ⁵√(−32) = −2 because (−2)⁵ = −32.
The product of an odd number of negative factors is negative, so one real root exists.

Index type	Formula	Negative radicand?	Example
Even (2, 4, 6...)	ⁿ√(aⁿ) = \|a\|	Not real	⁴√16 = 2; ⁴√(−16) not real
Odd (3, 5, 7...)	ⁿ√(aⁿ) = a	Real	³√(−8) = −2

🛠️ Simplifying radicals

🛠️ Product and quotient rules

Product rule for radicals: ⁿ√(A · B) = ⁿ√A · ⁿ√B

Quotient rule for radicals: ⁿ√(A/B) = ⁿ√A / ⁿ√B (where B ≠ 0)

A radical is simplified if the radicand contains no factors that are perfect powers of the index.
Strategy: factor the radicand to find the largest perfect power matching the index.

🛠️ Simplifying square roots

Look for perfect square factors (1, 4, 9, 16, 25, 36, 49, 64, 81, 100...).
Example: √150 = √(2 · 3 · 5²) = √(2 · 3) · √(5²) = √6 · 5 = 5√6.
Verification: 5√6 ≈ 12.25 and √150 ≈ 12.25; also (5√6)² = 25 · 6 = 150 ✓

🛠️ Simplifying cube roots

Look for perfect cube factors (1, 8, 27, 64, 125, 216...).
Example: ³√160 = ³√(2³ · 2² · 5) = ³√(2³) · ³√(20) = 2 · ³√20 = 2³√20.
With negative radicands (odd index): ³√(−320) = ³√((−1)⁵ · 2⁵ · 10) = −2 · ⁵√10.

🛠️ Simplifying higher roots

Example with quotient rule: ³√(−8/64) = ³√(−8) / ³√64 = ³√((−2)³) / ³√(4³) = −2/4 = −1/2.
Example with 4th root: ⁴√(80/81) = ⁴√80 / ⁴√81 = ⁴√(16 · 5) / 3 = 2⁴√5 / 3.
Don't confuse: use prime factorization to identify the largest perfect power systematically.

🛠️ Practical tip

If the radicand is not a perfect power, the root is irrational and can be approximated with a calculator.
Example: ³√2 ≈ 1.260 because 1.260³ ≈ 2.
Always give exact answers (like 5√6) unless an approximation is requested.

Note: The excerpt contains exercise problems and answers but focuses primarily on definitions, properties, and worked examples for roots and radicals. The key takeaway is understanding when to apply absolute values (even indices) versus when roots can be negative (odd indices), and how to simplify by factoring out perfect powers.

Simplifying Radical Expressions

🧭 Overview

🧠 One-sentence thesis

Simplifying radical expressions relies on identifying factors with powers that match the index and applying product and quotient rules to extract perfect powers from under the radical.

📌 Key points (3–5)

What simplification means: A radical is simplified when its radicand contains no factors that can be written as perfect powers of the index.
Core technique: Look for factors with powers matching the index (e.g., square factors for square roots, cubic factors for cube roots), then apply the property that the nth root of a to the nth power equals a (when a is nonnegative).
Variable assumption: Throughout this section, all variables are assumed positive to avoid absolute-value technicalities; without this assumption, even-index radicals require absolute values to ensure positive results.
Common confusion: When the index is even (like square roots), the result must be non-negative; when odd (like cube roots), negative radicands are allowed and produce negative results.
Why it matters: Simplified radicals are needed for formulas involving pendulum periods, free-fall time, vehicle speed estimation, and distance calculations.

🔍 Core simplification method

🔍 What makes a radical "simplified"

A radical expression is simplified if its radicand does not contain any factors that can be written as perfect powers of the index.

The goal is to move factors out from under the radical symbol whenever possible.
Example: The square root of 12 x to the 6th power y cubed can be simplified because 12 contains the square factor 4, x to the 6th is a perfect square (x cubed squared), and y cubed contains the square factor y squared.
What remains inside must have no perfect-power factors matching the index.

🧮 The key property for simplification

The excerpt uses the property: the nth root of (a to the nth power) equals a, provided a is nonnegative.

For odd indices (cube roots, fifth roots, etc.): the result equals a directly, even if a is negative.
For even indices (square roots, fourth roots, etc.): the result must be non-negative; if variables could be negative, use absolute value.
The excerpt's assumption: all variables represent positive numbers, so absolute values are not needed in the body examples.
Example: The cube root of 27 x cubed equals the cube root of (3 cubed times x cubed), which simplifies to 3x.

🔧 Step-by-step process

Identify perfect-power factors of the radicand that match the index.
- For square roots: look for square factors (4, 9, 16, x squared, y to the 4th, etc.).
- For cube roots: look for cubic factors (8, 27, x cubed, y to the 6th, etc.).
Rewrite the radicand as a product of these perfect-power factors and any remaining factors.
Apply the product rule for radicals: the nth root of (A times B) equals (nth root of A) times (nth root of B).
Simplify each perfect-power radical using the key property.
Multiply the simplified factors outside the radical; leave non-perfect-power factors inside.

Example: To simplify the square root of (18 a to the 5th over b to the 8th):

18 = 2 times 9 (9 is a square factor).
a to the 5th = (a squared) squared times a (a to the 4th is a perfect square).
b to the 8th = (b to the 4th) squared (perfect square).
Result: (3 a squared times square root of 2a) divided by b to the 4th.

🧩 Working with different indices

🧩 Square roots (index 2)

Look for factors that are perfect squares: 4, 9, 16, 25, x squared, y to the 4th, etc.
Example: The square root of (12 x to the 6th y cubed) becomes 2 x cubed y times the square root of (3y), because 12 = 4 times 3, x to the 6th = (x cubed) squared, and y cubed = y squared times y.
Don't confuse: x to the 6th is a perfect square (it's x cubed squared), but x to the 5th is not (it's x to the 4th times x, leaving one x inside).

🧊 Cube roots (index 3)

Look for factors that are perfect cubes: 8, 27, 64, 125, x cubed, y to the 6th, etc.
Example: The cube root of (80 x to the 5th y to the 7th) simplifies by recognizing 80 = 8 times 10 (8 is a perfect cube), x to the 5th = x cubed times x squared, and y to the 7th = (y squared) cubed times y.
Result: 2 x y squared times the cube root of (10 x squared y).

🔢 Higher indices (4th, 5th roots, etc.)

For fourth roots, look for factors that are 4th powers: 16, 81, x to the 4th, y to the 8th, etc.
For fifth roots, look for 5th powers: 32, 243, x to the 5th, y to the 10th, etc.
Helpful tip from the excerpt: Divide the exponent by the index. The quotient is the exponent outside the radical; the remainder is the exponent left inside.
- Example: For the cube root of (a to the 5th), divide 5 by 3 → quotient 1, remainder 2 → result is a times the cube root of (a squared).

🧮 Product and quotient rules

🧮 Product rule for radicals

The nth root of (A times B) equals (nth root of A) times (nth root of B).

This allows you to split a radical into separate factors.
Example: The square root of (4 times 10) equals (square root of 4) times (square root of 10) = 2 times square root of 10.
Use this to separate perfect-power factors from non-perfect-power factors.

➗ Quotient rule for radicals

The nth root of (A divided by B) equals (nth root of A) divided by (nth root of B).

Example: The square root of (9 over 25) equals (square root of 9) divided by (square root of 25) = 3 divided by 5.
This is used when simplifying fractions under a radical.

📐 Formulas involving radicals

⏱️ Pendulum period formula

The period T in seconds of a pendulum is given by: T equals 2 pi times the square root of (L divided by 32), where L is the length in feet.

Example from the excerpt: If L equals 3 halves (1.5 feet), substitute and simplify:
- T = 2 pi times square root of (3 over 64).
- Simplify: square root of 3 divided by 8.
- Result: (pi times square root of 3) divided by 4, approximately 1.36 seconds.

📏 Distance formula

Distance formula: Given two points (x₁, y₁) and (x₂, y₂), the distance d between them is the square root of [(x₂ − x₁) squared plus (y₂ − y₁) squared].

This formula comes from the Pythagorean theorem applied to a right triangle formed by the two points.
Example: Distance between (−5, 3) and (1, 1):
- Horizontal leg: 1 − (−5) = 6 units.
- Vertical leg: 3 − 1 = 2 units.
- Distance: square root of (36 + 4) = square root of 40 = 2 times square root of 10 units.

🔺 Pythagorean theorem applications

The Pythagorean theorem states: a squared plus b squared equals c squared, where c is the hypotenuse of a right triangle.

The hypotenuse equals the square root of (a squared plus b squared).
Example: To check if three points form a right triangle, calculate all three side lengths using the distance formula, then verify whether the sum of the squares of the two shorter sides equals the square of the longest side.
Don't confuse: The distance formula is derived from the Pythagorean theorem but is used for any two points, not just right triangles.

🚗 Vehicle speed estimation

The excerpt mentions a formula for estimating vehicle speed from skid marks: v equals 2 times the square root of (3d) on wet concrete, where d is skid mark length in feet and v is speed in miles per hour.

Example: If skid marks measure 27 feet, v = 2 times square root of 81 = 2 times 9 = 18 miles per hour.

⚠️ Important conventions and confusions

⚠️ The variable-positivity assumption

The excerpt explicitly states: "In this section, we will assume that all variables are positive."

Why this matters: Without this assumption, even-index radicals (square roots, fourth roots) require absolute values to ensure non-negative results.
Example: The square root of (x squared) equals the absolute value of x if x could be negative, but equals x if x is assumed positive.
The excerpt uses this assumption to "focus on calculating nth roots without the technicalities associated with the principal nth root problem."

⚠️ Even vs. odd indices

Index type	Behavior with negatives	Simplification rule
Even (2, 4, 6...)	Negative radicands not allowed (not real numbers)	Result must be non-negative; use absolute value if variables could be negative
Odd (3, 5, 7...)	Negative radicands allowed	Result can be negative; no absolute value needed

Example: The square root of (−1) is not a real number, but the cube root of (−1) equals −1.
Example: The cube root of (−32 x cubed y to the 6th z to the 5th) equals −2yz times the cube root of (x cubed y), with the negative sign preserved.

⚠️ Rewriting with coefficient 1

The excerpt includes exercises to "rewrite as a radical expression with coefficient 1."

This means moving the coefficient inside the radical by raising it to the index power.
Example: 3 times the square root of x equals the square root of (9x), because 3 squared equals 9.
Example: 2 times the cube root of (3x) equals the cube root of (8 times 3x) = cube root of (24x), because 2 cubed equals 8.

Adding and Subtracting Radical Expressions

🧭 Overview

🧠 One-sentence thesis

Radical expressions can be combined through addition and subtraction only when they are "like radicals" (sharing the same index and radicand), and simplification is often necessary before identifying which terms can be combined.

📌 Key points (3–5)

Like radicals definition: Radicals are "like" or "similar" when they share both the same index and the same radicand.
Combining rule: Add or subtract only the coefficients of like radicals; the radical part stays unchanged.
Simplification first: Often you must simplify each radical completely before you can tell which terms are like radicals.
Common confusion: The square root of a sum is NOT the sum of square roots—√(a + b) ≠ √a + √b; the same applies to differences and other indices.
Distributive property applies: When subtracting expressions with multiple terms, distribute the negative sign before combining like radicals.

🔍 What are like radicals?

🔍 Definition and criteria

Like radicals (or similar radicals): Radicals that share the same index and radicand.

Both the index (the small number indicating the root type) and the radicand (the expression under the radical) must match exactly.
Example: 2√6 and 5√6 are like radicals because both have index 2 (square root) and radicand 6.
Example: ³√(5x) and ³√(5x) are like radicals; ³√(5x) and ³√(2x) are NOT like radicals (different radicands).

🔍 When radicals are NOT like

Different radicands: √5 and √2 cannot be combined.
Different indices: ³√10 and √10 cannot be combined (one is a cube root, the other a square root).
Don't confuse: Even if the numerical values are close, radicals must have identical index and radicand to be like.

➕ Combining like radicals

➕ The basic rule

When adding or subtracting like radicals, add or subtract only the coefficients; the radical part remains the same.

The distributive property justifies this: 2√6 + 5√6 = (2 + 5)√6 = 7√6.
Typically, the distributive step is not shown; you simply combine coefficients directly.

➕ Addition example

Example: 7³√5 + 3³√5

Both terms share index 3 and radicand 5.
Combine coefficients: (7 + 3)³√5 = 10³√5.

➖ Subtraction example

Example: 4√10 − 5√10

Combine coefficients: (4 − 5)√10 = −1√10 = −√10.

➕ Multiple terms with different radicals

Example: 10√5 + 6√2 − 9√5 − 7√2

Group like radicals: (10√5 − 9√5) + (6√2 − 7√2) = √5 − √2.
Cannot simplify further because √5 and √2 are not like radicals.

🔧 Simplifying before combining

🔧 Why simplification matters

At first glance, radicals may not appear to be like radicals, but after simplifying each term completely, you may discover they share the same index and radicand.

🔧 Simplification process

Example: √32 − √18 + √50

Factor each radicand to find perfect squares:
- √32 = √(16·2) = 4√2
- √18 = √(9·2) = 3√2
- √50 = √(25·2) = 5√2
Now all terms are like radicals: 4√2 − 3√2 + 5√2 = 6√2.

🔧 Cube roots and higher indices

Example: ³√108 + ³√24 − ³√32 − ³√81

Factor to find perfect cubes:
- ³√108 = ³√(27·4) = 3³√4
- ³√24 = ³√(8·3) = 2³√3
- ³√32 = ³√(8·4) = 2³√4
- ³√81 = ³√(27·3) = 3³√3
Group like radicals: (3³√4 − 2³√4) + (2³√3 − 3³√3) = ³√4 − ³√3.

🧮 Working with variables

🧮 Assumption

In this section, all radicands containing variable expressions are assumed to be nonnegative.

🧮 Combining variable radicals

Example: −9³√(5x) − ³√(2x) + 10³√(5x)

Combine like radicals: (−9 + 10)³√(5x) − ³√(2x) = ³√(5x) − ³√(2x).
Cannot combine further because ³√(5x) and ³√(2x) have different radicands.

🧮 Distributing negatives

When subtracting an expression with multiple radical terms, apply the distributive property first.

Example: (5√x − 4√y) − (4√x − 7√y)

Distribute the negative: 5√x − 4√y − 4√x + 7√y.
Group like radicals: (5√x − 4√x) + (−4√y + 7√y) = √x + 3√y.

🧮 Complex variable expressions

Example: 5³√(3x⁴) + ³√(24x³) − (x·³√(24x) + 4³√(3x³))

Step 1: Simplify each radical by factoring perfect cubes.
Step 2: Distribute any negatives.
Step 3: Combine all like radicals by adding/subtracting coefficients.

⚠️ Critical warning: sums vs. products

⚠️ The key distinction

Operation	Under the radical	Property exists?
Product	√(x²y²) = xy	✓ Yes: √(a·b) = √a · √b
Sum	√(x² + y²)	✗ No: √(a + b) ≠ √a + √b

The property for splitting radicals applies only to multiplication, not addition or subtraction.
Example of what NOT to do: √5 − √2 ≠ √(5 − 2) = √3. (You can verify with a calculator: √5 − √2 ≈ 0.82, but √3 ≈ 1.73.)

⚠️ General form

For any index n:

ⁿ√a ± ⁿ√b ≠ ⁿ√(a ± b).

📐 Application: perimeter problems

📐 Using the distance formula

When calculating perimeters of triangles with given vertices, use the distance formula for each side, then simplify and combine like radicals.

Example: Triangle with vertices (−2, −1), (−3, 6), and (2, 1).

Calculate each side length using the distance formula.
Simplify each radical (e.g., √50 = 5√2, √20 = 2√5).
Add the three side lengths, combining like radicals where possible.
Final perimeter: 10√2 + 2√5 units.

📐 General steps

Use distance formula: √[(x₂ − x₁)² + (y₂ − y₁)²] for each side.
Simplify each resulting radical.
Combine like radicals to express the perimeter in simplest form.

Multiplying and Dividing Radical Expressions

🧭 Overview

🧠 One-sentence thesis

Multiplying and dividing radical expressions relies on product and quotient rules, and rationalizing the denominator ensures that final answers contain no radicals in the denominator.

📌 Key points (3–5)

Product rule for radicals: When multiplying radicals with the same index, multiply coefficients and radicands separately, then simplify.
Quotient rule for radicals: When dividing radicals with the same index, divide radicands and simplify the result.
Rationalizing the denominator: Find an equivalent expression with a rational (non-radical) denominator by multiplying by a strategic form of 1.
Common confusion: Conjugates eliminate radicals in denominators with two terms, but single-term denominators require multiplying by the nth root of missing factors.
Why it matters: Standard practice in mathematics is to express radical expressions without radicals in the denominator.

🔢 Multiplying radical expressions

🔢 Product rule for radicals

Product rule: Given real numbers with the same index n, the nth root of A times the nth root of B equals the nth root of (A times B).

Multiply coefficients together and radicands together.
After multiplying, simplify the result if possible.
Example: The cube root of 12 times the cube root of 6 equals the cube root of 72, which simplifies to 2 times the cube root of 9.

🔄 Multiplying coefficients and radicands

When coefficients appear in front of radicals, multiply them separately from the radicands.
Multiplication is commutative, so you can rearrange factors.
Example: 3 times the square root of 6, multiplied by 5 times the square root of 2, equals 15 times the square root of 12, which simplifies to 30 times the square root of 3.

📦 Distributive property with radicals

Apply the distributive property when one factor has multiple terms.
Multiply each term in the parentheses by the term outside.
Example: 5 times the square root of (2x), multiplied by (3 times the square root of x minus the square root of 2x), distributes to two separate products that are then simplified.

🔗 Multiplying binomials with radicals

Use the same process as multiplying polynomials: distribute each term.
Combine like radicals after multiplying.
Example: (square root of x minus 5 times the square root of y) squared expands to x minus 10 times the square root of xy plus 25y.

⚡ Conjugates

Conjugates: The binomials (a + b) and (a − b) are conjugates.

When multiplying conjugate binomials, the middle terms cancel out.
The result is always a rational expression: (square root of x plus square root of y) times (square root of x minus square root of y) equals x minus y.
This property is crucial for rationalizing denominators with two terms.
Example: (square root of 10 plus square root of 3) times (square root of 10 minus square root of 3) equals 10 minus 3, which equals 7.

➗ Dividing radical expressions

➗ Quotient rule for radicals

Quotient rule: Given real numbers with the same index n, the nth root of A divided by the nth root of B equals the nth root of (A divided by B).

Write the division as a single radical when possible.
Cancel common factors in the radicand before simplifying.
Example: The cube root of 96 divided by the cube root of 6 equals the cube root of 16, which simplifies to 2 times the cube root of 2.

🧮 Simplifying before dividing

Look for common factors in numerator and denominator radicands.
Reduce the fractional radicand before taking the root.
Example: The square root of (50x⁶y⁴) divided by the square root of (8x³y) simplifies by canceling to (5xy times the square root of xy) divided by 2.

🔄 Rationalizing the denominator

🎯 What rationalizing means

Rationalizing the denominator: The process of finding an equivalent expression with a rational (non-radical) denominator.

Standard mathematical practice avoids radicals in denominators.
Multiply by a special form of 1 so the denominator becomes rational.
The numerator and denominator must be multiplied by the exact same nonzero factor.

🔢 Single-term denominators with square roots

Multiply numerator and denominator by the same radical that appears in the denominator.
The denominator becomes a perfect square.
Example: 2 divided by the square root of (5x) is multiplied by (square root of 5x) over (square root of 5x), yielding (2 times the square root of 5x) divided by 5x.
Don't confuse: Factors inside a radical cannot cancel with factors outside.

🧊 Single-term denominators with higher roots

For cube roots and higher, multiply by the nth root of factors that produce nth powers.
Identify what power is needed to match the index.
Example: To rationalize 1 divided by the cube root of x, multiply by (cube root of x²) over (cube root of x²), because x times x² equals x³.
Example: 2 divided by the cube root of 25 requires multiplying by (cube root of 5) over (cube root of 5), because 25 times 5 equals 125, which is 5³.

🔗 Two-term denominators (conjugates)

When the denominator has two terms involving square roots, multiply by the conjugate.
The conjugate has the same terms but the opposite sign between them.
Multiplying by the conjugate eliminates the radicals in the denominator.
Example: 1 divided by (square root of 5 minus square root of 3) is multiplied by (square root of 5 plus square root of 3) over (square root of 5 plus square root of 3), yielding (square root of 5 plus square root of 3) divided by 2.

📐 Using the difference of squares formula

The product of conjugates follows the pattern: (a + b)(a − b) = a² − b².
For radicals: (square root of x plus square root of y) times (square root of x minus square root of y) equals x minus y.
This shortcut speeds up the multiplication step.
Example: (square root of 10 plus square root of 3) times (square root of 10 minus square root of 3) equals 10 minus 3, which equals 7.

⚠️ Common pitfalls

After rationalizing, you may need to reduce or cancel common factors.
Variables inside radicals do not cancel with variables outside radicals.
Always simplify the numerator after multiplying by the rationalizing factor.
Example: After rationalizing, (3a times the square root of 12ab) divided by (6ab) simplifies by factoring and canceling to yield (3 times the square root of 3ab) divided by b.

Rational Exponents

🧭 Overview

🧠 One-sentence thesis

Rational (fractional) exponents provide an alternative notation for radicals, where the denominator indicates the root index and the numerator indicates the power, and all standard exponent rules apply to these expressions.

📌 Key points (3–5)

What rational exponents represent: A fractional exponent m/n means the nth root of a raised to the m power.
Conversion between forms: Any radical can be written with rational exponents (exponential form) and vice versa (radical form).
Order flexibility: Taking the nth root first or applying the power first yields the same result.
Common confusion: Different indices—radicals with different indices cannot be combined directly; convert to exponential form first.
Why it matters: Rational exponents allow use of all exponent rules for simplifying and performing operations on radical expressions.

🔄 Converting between radical and exponential forms

🔄 Basic fractional exponents with numerator 1

a^(1/n) = nth root of a

The denominator of the fractional exponent determines the index of the root.
Example: 5^(1/2) equals the square root of 5 because two equal factors of 5^(1/2) multiply to give 5.
Example: 2^(1/3) equals the cube root of 2 because three equal factors of 2^(1/3) multiply to give 2.

🔢 General rational exponents

a^(m/n) = nth root of (a^m)

The denominator n is the index of the root.
The numerator m is the exponent applied to the base.
Example: 6^(2/5) can be written as the 5th root of (6 squared), which equals the 5th root of 36.
Radical form: expressions written with radical symbols.
Exponential form: expressions written with rational exponents.

⚖️ Two equivalent approaches

The excerpt shows that a^(m/n) can be evaluated in two ways:

Apply the power first: a^(m/n) = nth root of (a^m)
Apply the root first: a^(m/n) = (nth root of a)^m

Example: 27^(2/3) can be computed as:

Cube root of (27 squared) = cube root of 729 = 9, OR
(Cube root of 27) squared = 3 squared = 9

Both methods give the same answer; choose whichever makes the arithmetic simpler.

🧮 Simplifying expressions with rational exponents

🧮 Working with prime factorization

When simplifying expressions like 27^(2/3):

Replace the base with its prime factorization: 27 = 3^3
Apply exponent rules: (3^3)^(2/3) = 3^(3·2/3) = 3^2 = 9
This approach avoids working with very large numbers.

🔍 Reducing fractional exponents

When converting radicals to exponential form, simplify the fraction:

Example: The 6th root of (y cubed) = y^(3/6) = y^(1/2)
Always reduce the fractional exponent to lowest terms.

⚠️ Handling negative bases

Example: (-8)^(2/3)

First take the cube root: cube root of (-8) = -2
Then square: (-2)^2 = 4
Don't confuse: The order matters when dealing with negative bases and even powers.

📐 Operations using exponent rules

📐 Product and quotient rules

All standard exponent rules extend to rational exponents:

Rule	Formula	Example
Product rule	x^m · x^n = x^(m+n)	7^(1/3) · 7^(4/9) = 7^(7/9)
Quotient rule	x^m / x^n = x^(m-n)	x^(3/2) / x^(2/3) = x^(5/6)
Power rule	(x^m)^n = x^(m·n)	(y^(3/4))^(2/3) = y^(1/2)

When adding or subtracting exponents, find a common denominator.
Example: 1/3 + 4/9 = 3/9 + 4/9 = 7/9

🔨 Power rules for products and quotients

Power of a product: (xy)^n = x^n · y^n
Power of a quotient: (x/y)^n = x^n / y^n

Example: (81a^8 b^12)^(3/4)

Apply the power to each factor: 81^(3/4) · (a^8)^(3/4) · (b^12)^(3/4)
Simplify: 27a^6 b^9

➖ Negative and zero exponents

Negative exponent: x^(-n) = 1 / x^n
Zero exponent: x^0 = 1 (when x ≠ 0)

Example: (9x^4)^(-3/2) = 1 / (9x^4)^(3/2) = 1 / (27x^6)

🔀 Radicals with different indices

🔀 Why conversion is necessary

The product and quotient rules for radicals only work when the indices are the same.
If indices differ, convert to exponential form first.

🔀 Multiplying radicals with different indices

Example: square root of 2 times cube root of 2

Convert: 2^(1/2) · 2^(1/3)
Find common denominator: 2^(3/6) · 2^(2/6) = 2^(5/6)
Result: 6th root of (2^5)

🔀 Dividing radicals with different indices

Example: cube root of 4 divided by 5th root of 2

Convert: 2^(2/3) / 2^(1/5)
Subtract exponents: 2^(2/3 - 1/5) = 2^(7/15)
Result: 15th root of (2^7)

🔀 Nested radicals

Example: square root of (cube root of 4)

Convert the inner radical first: cube root of 4 = 4^(1/3) = 2^(2/3)
Apply the outer radical: (2^(2/3))^(1/2) = 2^(1/3)
Result: cube root of 2
This uses the power rule: multiply the exponents (2/3)·(1/2) = 1/3.

Solving Radical Equations

🧭 Overview

🧠 One-sentence thesis

Solving radical equations requires isolating the radical, raising both sides to the appropriate power to eliminate it, and then checking all solutions because the squaring (or power) process can introduce extraneous solutions that do not satisfy the original equation.

📌 Key points (3–5)

What a radical equation is: any equation containing one or more radicals with a variable inside the radicand.
Core solving method: isolate the radical term, then raise both sides to the power matching the index (square for square roots, cube for cube roots, etc.) to eliminate the radical.
Why checking is mandatory: squaring (or raising to any power) both sides can create extraneous solutions—values that solve the transformed equation but not the original.
Common confusion: the converse of the squaring property is not true; if a² = b², it does not guarantee a = b (e.g., (−3)² = 3² but −3 ≠ 3).
Multiple radicals: when more than one radical appears, isolate and eliminate them one at a time, applying the power property repeatedly.

🔑 What are radical equations and why squaring works

🔑 Definition and examples

Radical equation: any equation that contains one or more radicals with a variable in the radicand.

The radicand is the expression inside the radical symbol.
Examples from the excerpt:
- √(2x − 1) = 3
- ∛(4x² + 7) − 2 = 0
- √(x + 2) − √x = 1

⚙️ The squaring property of equality

Squaring property of equality: Given real numbers a and b, if a = b, then a² = b².

This property allows us to eliminate square roots by squaring both sides.
Example: if √x = 5, then (√x)² = 5², so x = 25.
The property extends to any positive integer power n: if a = b, then aⁿ = bⁿ (the power property of equality).

⚠️ Why the converse fails

The converse—if a² = b², then a = b—is not necessarily true.
Example: (−3)² = 3² gives 9 = 9, but −3 ≠ 3.
This asymmetry is why extraneous solutions can appear.

🛠️ Step-by-step solving process

🛠️ Four-step method (for square roots)

The excerpt outlines a standard procedure:

Isolate the square root: move all other terms to the opposite side so the radical stands alone.
Square both sides: apply (√...)² = ... to eliminate the radical.
Solve the resulting equation: this will be a polynomial (linear, quadratic, etc.).
Check all solutions in the original equation: discard any extraneous solutions.

📐 Example walkthrough: √(3x + 1) = 4

Step 1: The radical is already isolated.
Step 2: Square both sides: (√(3x + 1))² = 4² → 3x + 1 = 16.
Step 3: Solve: 3x = 15 → x = 5.
Step 4: Check: √(3·5 + 1) = √16 = 4 ✓.
Answer: x = 5.

🔄 When both sides have radicals

If radicals appear on both sides (e.g., √(5x − 3) = √(4x − 1)), you can still square both sides directly.
Example: (√(5x − 3))² = (√(4x − 1))² → 5x − 3 = 4x − 1 → x = 2.
Always check: √(5·2 − 3) = √7 and √(4·2 − 1) = √7 ✓.

🚨 Extraneous solutions and checking

🚨 What extraneous solutions are

Extraneous solution: a properly found solution that does not solve the original equation.

Squaring can introduce solutions that satisfy the squared equation but not the original.
Example: √x = −5 has no real solution, but squaring gives x = 25, which does not check.

🔍 Why checking is not optional

The excerpt emphasizes: "Checking the solutions after squaring both sides of an equation is not optional."
Use the original equation when checking, not the transformed one.
Example from the excerpt: solving √(x − 3) = x − 5 yields x = 4 and x = 7.
- Check x = 4: √(4 − 3) = 1 but 4 − 5 = −1 → 1 ≠ −1 ✗ (extraneous).
- Check x = 7: √(7 − 3) = 2 and 7 − 5 = 2 ✓.
- Only x = 7 is valid.

🛑 When all solutions are extraneous

Sometimes every candidate fails the check, leaving no solution (denoted Ø).
Example: √(4 − 11x) − x + 2 = 0 produces x = 0 and x = −7, but both fail the check.

🔢 Solving equations with higher-index radicals

🔢 Cube roots and beyond

For cube roots (∛), cube both sides: (∛...)³ = ....
For nth roots (ⁿ√), raise both sides to the nth power.
Example: ∛(4x² + 7) − 2 = 0
- Isolate: ∛(4x² + 7) = 2.
- Cube: (∛(4x² + 7))³ = 2³ → 4x² + 7 = 8 → 4x² = 1 → x = ±1/2.
- Check both: both x = 1/2 and x = −1/2 satisfy the original equation ✓.

🔄 Checking is still required

Even though cubing (or higher powers) can also introduce extraneous solutions, the excerpt shows that checking remains essential.
The power property of equality applies to any positive integer n, but the converse still does not hold in general.

🔗 Equations with multiple radical terms

🔗 Isolate one radical at a time

When an equation has more than one radical, isolate one, eliminate it, then repeat for the next.
Example: √(x + 2) − √x = 1
- Isolate √(x + 2): √(x + 2) = √x + 1.
- Square: x + 2 = (√x + 1)² = x + 2√x + 1.
- Simplify: 1 = 2√x.
- Isolate again: √x = 1/2.
- Square again: x = 1/4.
- Check: √(1/4 + 2) − √(1/4) = √(9/4) − 1/2 = 3/2 − 1/2 = 1 ✓.

⚠️ Common mistake: squaring each term separately

Incorrect: (√(x + 2))² − (√x)² = 1² → x + 2 − x = 1 → 2 = 1 ✗.
Why it's wrong: (A + B)² ≠ A² + B²; you must apply the distributive property (FOIL).
The excerpt explicitly warns: "This is a common mistake and leads to an incorrect result."

📐 Example with two radicals: √(2x + 10) − √(x + 6) = 1

Isolate √(2x + 10): √(2x + 10) = √(x + 6) + 1.
Square: 2x + 10 = (√(x + 6) + 1)² = x + 6 + 2√(x + 6) + 1 = x + 7 + 2√(x + 6).
Simplify: x + 3 = 2√(x + 6).
Square again: (x + 3)² = 4(x + 6) → x² + 6x + 9 = 4x + 24 → x² + 2x − 15 = 0.
Factor: (x − 3)(x + 5) = 0 → x = 3 or x = −5.
Check x = 3: √16 − √9 = 4 − 3 = 1 ✓.
Check x = −5: √0 − √1 = 0 − 1 = −1 ≠ 1 ✗.
Answer: x = 3.

📊 Summary table: key distinctions

Aspect	What to do	What NOT to do
Isolating	Move all non-radical terms away from the radical	Leave multiple terms on the same side as the radical without isolating
Squaring	Square the entire side as a binomial/trinomial (use distributive property)	Square each term separately (e.g., (A + B)² ≠ A² + B²)
Checking	Substitute back into the original equation	Skip checking or use the transformed equation
Extraneous solutions	Discard any solution that fails the check	Assume all algebraic solutions are valid
Multiple radicals	Isolate and eliminate one at a time, squaring repeatedly	Try to square all radicals at once

🎯 Geometric interpretation

🎯 Graphical view of solutions

The excerpt notes that solving √(3x + 1) = 4 is equivalent to finding where f(x) = √(3x + 1) intersects g(x) = 4.
The intersection point x = 5 is the solution.
Similarly, √(x − 3) = x − 5 corresponds to where f(x) = √(x − 3) meets g(x) = x − 5, which occurs at x = 7 (not x = 4, which is extraneous).

📈 Why extraneous solutions appear graphically

Squaring can "reflect" part of a graph, creating intersection points that do not correspond to the original equation.
Checking ensures we only accept points that lie on both original curves.

🔧 Special cases and tips

🔧 Equations with coefficients on the radical

A term like 2√(2x + 5) is still considered isolated (terms are separated by + or −, not by coefficients).
Example: 2√(2x + 5) − x = 4 → isolate: 2√(2x + 5) = x + 4 → square: 4(2x + 5) = (x + 4)².

🔧 Rational exponents

The excerpt mentions that x^(1/2) is equivalent to √x, and x^(1/3) to ∛x.
Solving x^(1/2) − 10 = 0 is the same as solving √x = 10 → x = 100.

🔧 Solving for other variables

The same techniques apply when solving for a variable other than x.
Example: given r = √P − 1, solve for P:
- Isolate √P: √P = r + 1.
- Square: P = (r + 1)².

📝 Key takeaways for practice

Always isolate before powering: move all non-radical terms away from the radical.
Apply the power matching the index: square for √, cube for ∛, etc.
Expand carefully: use the distributive property (FOIL) when squaring binomials; never square term-by-term.
Check every solution: substitute back into the original equation and discard any that do not work.
Expect extraneous solutions: they are a natural consequence of the squaring/power process, not a sign of error in your algebra.
For multiple radicals: isolate and eliminate one at a time, repeating the process until only a polynomial remains.

Complex Numbers and Their Operations

🧭 Overview

🧠 One-sentence thesis

Complex numbers extend the real number system by introducing the imaginary unit i to represent square roots of negative numbers, and they follow consistent algebraic rules for addition, subtraction, multiplication, and division.

📌 Key points (3–5)

The imaginary unit: i is defined as the square root of −1, where i squared equals −1.
Standard form: Complex numbers are written as a + bi where a is the real part and b is the imaginary part; real numbers are a subset of complex numbers.
Operations preserve structure: Adding, subtracting, multiplying, and dividing complex numbers always produces another complex number.
Common confusion: The product and quotient rules for radicals only work when both radicands are positive; always convert to imaginary unit form before multiplying square roots of negative numbers.
Complex conjugates: Multiplying a complex number by its conjugate (a + bi and a − bi) produces a real number, which is the key technique for division.

🔢 The imaginary unit and complex number definition

🔢 What is the imaginary unit i

The imaginary unit i is defined as the square root of −1, where i squared equals −1.

Before this definition, square roots of negative numbers were undefined in the real number system.
The imaginary unit resolves this by allowing us to express square roots of negative numbers.
For any non-negative real number a: the square root of negative a equals i times the square root of a.
Example: The square root of −9 equals 3i, and when you square 3i you get 9 times i squared, which equals 9 times (−1), which equals −9.

📐 Standard form of complex numbers

A complex number is any number of the form a + bi, where a and b are real numbers.

The real part is a and the imaginary part is b (note: b itself is real, but it's the coefficient of i).
Example: In 3 − 4i, the real part is 3 and the imaginary part is −4.
Real numbers are complex numbers with imaginary part zero: 5 can be written as 5 + 0i.
The set of real numbers is a subset of the set of complex numbers.

⚠️ Converting square roots of negatives

When writing square roots of negative numbers in terms of i, place i in front of the radical to avoid confusion.

Correct notation: The square root of −7 equals i times the square root of 7.
Why this matters: Writing it as "the square root of 7 times i" might make i look like it's under the radical.
Example: The square root of −72 equals the square root of (−1 times 36 times 2), which equals i times 6 times the square root of 2, written as 6i times the square root of 2.

➕ Adding and subtracting complex numbers

➕ How to add complex numbers

Adding complex numbers works like combining like terms in polynomials: add the real parts together and add the imaginary parts together.

Given (a + bi) + (c + di), the result is (a + c) + (b + d)i.
Example: (5 − 2i) + (7 + 3i) = (5 + 7) + (−2 + 3)i = 12 + i.
The real parts (5 and 7) combine to 12; the imaginary parts (−2 and 3) combine to 1.

➖ How to subtract complex numbers

Subtracting complex numbers requires distributing the negative sign, then combining like terms.

Given (a + bi) − (c + di), the result is (a − c) + (b − d)i.
Example: (10 − 7i) − (9 + 5i) = 10 − 7i − 9 − 5i = (10 − 9) + (−7 − 5)i = 1 − 12i.
Don't confuse: The subtraction applies to both the real and imaginary parts of the second number.

🔄 Result of addition and subtraction

The sum or difference of complex numbers is always another complex number in standard form a + bi.

✖️ Multiplying complex numbers

✖️ Distributive property and i squared

Multiplying complex numbers uses the distributive property, combined with the fact that i squared equals −1.

Example: −6i times (2 − 3i) = (−6i times 2) − (−6i times 3i) = −12i + 18i squared.
Substitute i squared = −1: −12i + 18(−1) = −12i − 18 = −18 − 12i.
Always simplify by replacing i squared with −1.

🔢 Multiplying binomials

When multiplying two complex numbers in binomial form, distribute all terms and simplify.

Example: (3 − 4i)(4 + 5i) = 12 + 15i − 16i − 20i squared.
Combine like terms: 12 + 15i − 16i − 20(−1) = 12 − i + 20 = 32 − i.
General formula: (a + bi)(c + di) = (ac − bd) + (ad + bc)i.

🔗 Complex conjugates

Two complex numbers whose real parts are the same and imaginary parts are opposite are called complex conjugates.

If given a + bi, its complex conjugate is a − bi.
Key property: The product of complex conjugates is a real number: (a + bi)(a − bi) = a squared + b squared.
Example: (5 + 2i)(5 − 2i) = 25 − 10i + 10i − 4i squared = 25 − 4(−1) = 25 + 4 = 29.
This property is essential for division.

⚠️ Product rule caution

The product rule for radicals requires both radicands to be positive; it fails for negative numbers.

Incorrect: The square root of −4 times the square root of −9 does NOT equal the square root of 36.
Why it fails: The square root of −4 equals 2i and the square root of −9 equals 3i, so the product is 6i squared = −6, not 6.
Always convert to imaginary unit form first: The square root of −6 times the square root of −15 = (i times the square root of 6) times (i times the square root of 15) = i squared times the square root of 90 = −1 times 3 times the square root of 10 = −3 times the square root of 10.

➗ Dividing complex numbers

➗ Rationalization technique

To divide complex numbers, multiply both numerator and denominator by the complex conjugate of the denominator.

This technique is similar to rationalizing denominators with radicals.
The conjugate of the denominator makes the denominator real (using the product-of-conjugates property).
Example: To divide 1 by (2 − 3i), multiply by (2 + 3i) over (2 + 3i).

🔢 Step-by-step division

Example: 1 divided by (2 − 3i)

Identify the conjugate: 2 + 3i
Multiply: [1 times (2 + 3i)] divided by [(2 − 3i) times (2 + 3i)]
Simplify numerator: 2 + 3i
Simplify denominator: 4 + 9 = 13 (using the conjugate product formula)
Result: (2 + 3i) divided by 13 = 2/13 + (3/13)i

📐 General division formula

For (a + bi) divided by (c + di) where c and d are not both zero:

Multiply by (c − di) over (c − di)
Numerator becomes: (ac + bd) + (bc − ad)i
Denominator becomes: c squared + d squared (a real number)
Final form: [(ac + bd) divided by (c squared + d squared)] + [(bc − ad) divided by (c squared + d squared)]i

💡 Monomial denominators

When the denominator is a pure imaginary number (like 2i), you can multiply by i over i as a shortcut.

Example: (8 − 3i) divided by 2i can be multiplied by i over i.
Result: (8i − 3i squared) divided by (2i squared) = (8i + 3) divided by (−2) = −3/2 − 4i.
This avoids extra steps but achieves the same result as using the conjugate.

🎯 Key takeaways

🎯 Closure property

All operations (addition, subtraction, multiplication, division) on complex numbers produce complex numbers.

This means the set of complex numbers is "closed" under these operations.
Division is defined for all complex numbers except division by zero.

🎯 Why complex conjugates matter

Complex conjugates are the key tool for division because their product is always real.

The product (a + bi)(a − bi) = a squared + b squared contains no imaginary part.
This allows us to eliminate i from denominators.
Don't confuse: The conjugate changes the sign of the imaginary part only, not the real part.

🎯 Order of operations

When working with square roots of negative numbers, always convert to imaginary unit form before performing operations.

This prevents errors from misapplying the product rule for radicals.
Example workflow: See negative under radical → factor out −1 → write as i times the square root of the positive part → then multiply or divide.

Extracting Square Roots and Completing the Square

🧭 Overview

🧠 One-sentence thesis

Quadratic equations can be solved by extracting square roots when in the form ax² + c = 0, or by completing the square to transform any quadratic into a form where roots can be extracted.

📌 Key points (3–5)

Extracting roots: applies to equations of the form ax² + c = 0 (no middle term); isolate x² and apply the square root property x² = k → x = ±√k.
Completing the square: transforms any quadratic x² + bx + c = 0 into (x − p)² = q by adding (b/2)² to both sides, enabling root extraction.
Leading coefficient requirement: completing the square requires dividing by the leading coefficient first if it is not 1.
Common confusion: remember the ± symbol when extracting roots; both positive and negative roots are solutions.
Complex solutions: when extracting roots of negative numbers, solutions involve the imaginary unit i.

🔧 Extracting square roots method

🔧 What the square root property says

Square root property: For any real number k, if x² = k, then x = ±√k.

This property is the foundation for extracting roots.
The ± symbol is essential—it captures both the positive and negative square roots.
Example: If x² = 9, then x = ±3, giving solutions x = 3 and x = −3.

🎯 When to use extracting roots

Use this method when the quadratic has no middle term (b = 0), giving the form ax² + c = 0.
The quadratic expression does not need to factor for this method to work.
Example: 9x² − 8 = 0 can be solved by isolating x² to get x² = 8/9, then x = ±√(8/9) = ±(2√2)/3.

🔢 Steps for extracting roots

Isolate the squared term: Move the constant to the other side to get x² = (some value).
Apply the square root property: Take the square root of both sides, remembering ±.
Simplify: Simplify radicals and rationalize denominators if needed.
Separate solutions: The ± gives two separate equations to solve.

🌀 Complex solutions from negative values

When x² equals a negative number, there is no real solution.
Solutions are complex, using i = √(−1).
Example: x² + 25 = 0 gives x² = −25, so x = ±√(−25) = ±5i.

📦 Extracting roots from binomial squares

The method extends to equations like (x + 5)² = 9.
Apply the square root property directly: x + 5 = ±3.
Then solve: x = −5 + 3 = −2 or x = −5 − 3 = −8.
This is faster than expanding and factoring.

🧩 Completing the square method

🧩 What completing the square means

Completing the square: The process of rewriting a quadratic equation to be in the form (x − p)² = q.

This transforms any quadratic into a form where extracting roots is straightforward.
The key insight: any perfect square trinomial x² + bx + c has c = (b/2)².

🔍 How to find the completing value

For x² + bx + ?, the value that completes the square is (b/2)².
This comes from the pattern: (x + b/2)² = x² + bx + (b/2)².
Example: For x² − 6x + ?, use b = −6, so (−6/2)² = 9. Thus x² − 6x + 9 = (x − 3)².
Example: For x² + x + ?, use b = 1, so (1/2)² = 1/4. Thus x² + x + 1/4 = (x + 1/2)².

📋 Steps for completing the square (when leading coefficient is 1)

Rearrange to x² + bx = c: Move the constant term to the right side.
Calculate (b/2)²: Find the value that completes the square.
Add (b/2)² to both sides: This creates a perfect square trinomial on the left.
Factor the left side: Write as (x + b/2)².
Extract roots: Apply the square root property and solve.

⚠️ When the leading coefficient is not 1

Must divide first: If the equation is ax² + bx + c = 0 with a ≠ 1, divide every term by a before completing the square.
The formula (b/2)² only works when the coefficient of x² is exactly 1.
Example: For 2x² + 5x − 1 = 0, divide by 2 to get x² + (5/2)x − 1/2 = 0, then proceed.

🧮 Working with fractions

When b is odd or not divisible by 2, (b/2)² will be a fraction.
Example: For x² + 3x + 4 = 0, rearrange to x² + 3x = −4. Then (3/2)² = 9/4. Add 9/4 to both sides: x² + 3x + 9/4 = −4 + 9/4 = −7/4.
Factor: (x + 3/2)² = −7/4, so x = −3/2 ± √(−7/4) = −3/2 ± (√7/2)i.

🎓 Comparing the two methods

🎓 When each method is best

Method	Best for	Limitation
Extracting roots	Equations with no middle term (ax² + c = 0) or already in form (x − p)² = q	Only works when b = 0 or equation is already a binomial square
Completing the square	Any quadratic equation	More steps; can involve tedious fraction arithmetic

🔄 Why completing the square is universal

Extracting roots is faster but limited to special forms.
Completing the square works for any quadratic equation, even when it doesn't factor.
It's especially useful when solutions are irrational or complex.
Don't confuse: factoring only works when solutions are rational; completing the square always works.

💡 Key reminders

💡 Always include ±

The most common error is forgetting the ± when applying the square root property.
Every time you take the square root of both sides, you get two solutions (unless the right side is zero).

💡 Simplify and rationalize

After extracting roots, simplify all radical expressions.
Rationalize denominators: √(a/b) = √a/√b = (√a · √b)/(√b · √b) = (√(ab))/b.
Example: √(5/2) = √5/√2 = (√5 · √2)/2 = √10/2.

💡 Check for extraneous solutions

Although not emphasized in this excerpt, it's good practice to check solutions in the original equation.
The excerpt shows verification examples for several problems.

Quadratic Formula

🧭 Overview

🧠 One-sentence thesis

The quadratic formula provides a universal method to solve any quadratic equation in standard form, and the discriminant reveals the number and type of solutions without solving the equation.

📌 Key points (3–5)

What the quadratic formula does: solves any quadratic equation in standard form using the formula x = (negative b plus-or-minus square root of b squared minus 4ac) divided by 2a.
How to apply it: rewrite the equation in standard form, identify coefficients a, b, and c, substitute into the formula, and simplify.
What the discriminant reveals: the expression b² − 4ac determines whether solutions are real or complex, and whether they are rational or irrational.
Common confusion: discriminant value vs solution type—positive discriminant means two real solutions, zero means one real solution (double root), negative means two complex solutions.
Why it matters: the quadratic formula works when factoring fails, and the discriminant helps choose the best solving method.

📐 Deriving and using the formula

📐 Where the formula comes from

The excerpt derives the quadratic formula by completing the square on the general equation ax² + bx + c = 0:

Start with standard form and divide by a
Complete the square by adding (b/2a)² to both sides
Extract roots and simplify to get x = (−b ± √(b² − 4ac)) / 2a

Quadratic formula: x = (−b ± √(b² − 4ac)) / 2a, which gives solutions to any quadratic equation ax² + bx + c = 0 where a, b, c are real numbers and a ≠ 0.

🔢 Identifying coefficients

Before using the formula, identify a, b, and c from standard form ax² + bx + c = 0:

a is the coefficient of x²
b is the coefficient of x
c is the constant term
If a term is missing, use 0 as its coefficient

Example: For 3x² + 6x − 2 = 0, we have a = 3, b = 6, c = −2.

⚙️ Substitution and simplification

After identifying coefficients:

Substitute a, b, c into the formula
Simplify the expression under the square root (the radicand)
Simplify the square root if possible (factor out perfect squares)
Reduce the fraction by factoring common terms from numerator and denominator

Example: If the radicand is 60, rewrite as √(4 × 15) = 2√15 before simplifying the fraction.

🔍 Understanding the discriminant

🔍 What the discriminant is

Discriminant: the expression b² − 4ac inside the radical of the quadratic formula.

The discriminant determines solution characteristics without completing the full calculation:

It appears under the square root, so its sign affects whether solutions are real or complex
Its value as a perfect square affects whether solutions are rational or irrational

🎯 Three cases for solution types

Discriminant value	Number of solutions	Type of solutions
b² − 4ac > 0	Two real solutions	Rational if perfect square; irrational otherwise
b² − 4ac = 0	One real solution	Always rational (double root)
b² − 4ac < 0	Two complex solutions	Complex conjugate pairs

✅ Positive discriminant

When b² − 4ac > 0, the equation has two distinct real solutions.

If the discriminant is a perfect square (e.g., 49, 64), solutions are rational and the equation can be factored
If not a perfect square (e.g., 20, 41), solutions are irrational

Example: For x² − 2x − 4 = 0, the discriminant is (−2)² − 4(1)(−4) = 20, giving two irrational solutions 1 ± √5.

🔄 Zero discriminant

When b² − 4ac = 0, the ± term vanishes and the equation has exactly one solution (a double root).

Example: For 25x² − 20x + 4 = 0, the discriminant is (−20)² − 4(25)(4) = 0, giving the double root x = 2/5.

❌ Negative discriminant

When b² − 4ac < 0, the square root of a negative number introduces the imaginary unit i, producing two complex solutions.

Complex solutions always appear as conjugate pairs: a + bi and a − bi
The excerpt notes that irrational solutions also appear in conjugate pairs

Example: For x² − 4x + 29 = 0, the discriminant is (−4)² − 4(1)(29) = −100, giving solutions 2 ± 5i.

🛠️ Solving strategy

🛠️ Standard form first

Before applying the formula, rewrite the equation with zero on one side:

Expand any products
Combine like terms
Arrange as ax² + bx + c = 0

Example: For (5x + 1)(x − 1) = x(x + 1), expand both sides, collect terms to get 4x² − 5x − 1 = 0.

🔀 When to use the quadratic formula

The excerpt emphasizes choosing methods strategically:

Use factoring if the equation factors easily (discriminant is a perfect square)
Use extracting roots if the equation has form (expression)² = constant
Use the quadratic formula when factoring is difficult or impossible

Don't confuse: The quadratic formula always works, but other methods may be faster for special cases.

📝 Missing terms

When x or the constant term is missing, use 0 for that coefficient.

Example: For x² − 45 = 0, this is equivalent to 1x² + 0x − 45 = 0, so a = 1, b = 0, c = −45.

✔️ Checking and interpreting solutions

✔️ Verification

Substitute solutions back into the original equation to verify:

For complex solutions, remember that i² = −1
Both solutions from the ± should satisfy the equation

🎲 Real-world applications

The excerpt includes applications where the quadratic formula solves for:

Time (projectile motion, stopping distance)
Dimensions (rectangles, triangles)
Business quantities (profit, revenue)
Rates (travel speed, work rates)

In applied problems, round solutions appropriately and reject physically impossible values (e.g., negative time or dimensions).

Solving Equations Quadratic in Form

🧭 Overview

🧠 One-sentence thesis

Equations that are not themselves quadratic can often be transformed into quadratic form through substitution, allowing us to apply familiar quadratic-solving techniques to find all solutions.

📌 Key points (3–5)

What "quadratic in form" means: an equation of the form au² + bu + c = 0 where u represents an algebraic expression (not just x).
The u-substitution technique: replace a complex expression with a simpler variable u, solve the resulting quadratic, then back-substitute to find the original variable.
When to use each method: if c = 0, factor out the GCF; if b = 0, extract roots; otherwise check the discriminant to decide between factoring or the quadratic formula.
Common confusion: after solving for u, you must back-substitute and solve for the original variable—the u-values are not the final answer.
Multiple and complex roots: the fundamental theorem of algebra guarantees as many roots as the polynomial's degree when counting multiplicity and complex solutions.

🎯 General strategy for quadratic equations

🎯 Choosing the right method

The excerpt outlines a decision tree based on the coefficients of ax² + bx + c = 0:

If c = 0: Factor out the greatest common factor (GCF) and apply the zero-product property.
- Example: For 12x² − 3x = 0, factor to 3x(4x − 1) = 0, giving solutions x = 0 and x = 1/4.
If b = 0: Isolate x² and extract the square root.
- Example: For 5x² + 8 = 0, isolate x² = −8/5, then x = ± (square root of −8/5), which simplifies to ± 2i√10 / 5.
If a, b, and c are all nonzero: Calculate the discriminant b² − 4ac.
- If the discriminant is a perfect square, solve by factoring.
- If not a perfect square, use the quadratic formula.
- The discriminant tells you the nature of solutions: positive perfect square → two rational; positive non-square → two irrational; zero → one repeated; negative → two complex conjugates.

🎯 Why not always use the quadratic formula

The quadratic formula works for every quadratic equation, but factoring or extracting roots is often faster and less error-prone when applicable.

🔄 The u-substitution technique

🔄 What is quadratic form

Quadratic form: an equation of the form au² + bu + c = 0, where a, b, and c are real numbers and u represents an algebraic expression.

The key insight: if you can identify a repeated structure in an equation, you can treat that structure as a single variable.
This transforms a complicated equation into a familiar quadratic.

🔄 How u-substitution works

Step 1: Identify the substitution.

Look for an expression that appears both as itself and squared (or in two related powers).
Example: In x⁴ − 4x² − 32 = 0, notice x⁴ = (x²)². Let u = x².

Step 2: Rewrite the equation in terms of u.

Replace every instance: x⁴ becomes u², x² becomes u.
The equation becomes u² − 4u − 32 = 0.

Step 3: Solve the quadratic in u.

Use factoring, the quadratic formula, or another appropriate method.
Example: (u − 8)(u + 4) = 0 gives u = 8 or u = −4.

Step 4: Back-substitute and solve for the original variable.

Replace u with the original expression and solve.
Example: x² = 8 gives x = ±2√2; x² = −4 gives x = ±2i.
Don't confuse: the u-values are intermediate; you must complete the back-substitution.

Step 5: Check for extraneous solutions.

Especially important when the original equation involved radicals or even powers, as squaring can introduce false solutions.

🧮 Common substitution patterns

🧮 Polynomial powers

Fourth-degree: x⁴ − 4x² − 32 = 0 → let u = x², then u² − 4u − 32 = 0.
After solving for u, remember x² can equal a negative number (yielding complex x-values).

🧮 Radical expressions

Square roots: x − 2√x − 8 = 0 → let u = √x, then u² − 2u − 8 = 0.
After solving, square both sides to find x, then check all solutions (extraneous solutions are common).
Example: u = 4 gives x = 16 (valid); u = −2 gives x = 4 (extraneous, because √4 ≠ −2).

🧮 Fractional exponents

Rational exponents: x^(2/3) − 3x^(1/3) − 10 = 0 → let u = x^(1/3), then u² − 3u − 10 = 0.
Solve for u, then cube both sides to find x.
Example: u = 5 gives x = 125; u = −2 gives x = −8 (both valid because cube roots of negative numbers are real).

🧮 Negative exponents

Reciprocals: 3y^(−2) + 7y^(−1) − 6 = 0 → let u = y^(−1) = 1/y, then 3u² + 7u − 6 = 0.
Solve for u, then take reciprocals to find y.
Remember the original equation is undefined where y = 0, so check that solutions are not restrictions.

🧮 Compound expressions

Nested fractions or sums: ((t + 2)/t)² + 8((t + 2)/t) + 7 = 0 → let u = (t + 2)/t.
Solve the quadratic in u, then solve the resulting rational equations for t.

🔍 Checking and interpreting solutions

🔍 When to check solutions

Always check when the original equation involved:
- Radicals (even roots can introduce extraneous solutions).
- Even powers (squaring both sides is not reversible).
- Rational expressions (solutions might make denominators zero).

🔍 Counting all roots

Fundamental theorem of algebra: If multiple roots and complex roots are counted, every polynomial with one variable has as many roots as its degree.

A cubic (degree 3) has three roots (counting multiplicity and complex roots).
Example: x³ − 8 = 0 can be factored as (x − 2)(x² + 2x + 4) = 0, giving one real root (x = 2) and two complex roots (x = −1 ± i√3).
Don't confuse: "three roots" does not mean three distinct real roots; complex and repeated roots count toward the total.

🔍 Multiple roots

A root occurs multiple times if the same factor appears repeatedly.
Example: (x − 2)³ = 0 has degree 3 but only one distinct root, x = 2, occurring three times.

🧪 Non-factorable cases

🧪 Using the quadratic formula after substitution

When the quadratic in u does not factor, use the quadratic formula:

Example: x⁴ − 10x² + 23 = 0 → let u = x², giving u² − 10u + 23 = 0.
Apply the formula: u = (10 ± √(100 − 92)) / 2 = (10 ± √8) / 2 = 5 ± √2.
Back-substitute: x² = 5 − √2 gives x ≈ ±1.89; x² = 5 + √2 gives x ≈ ±2.53.
The excerpt notes that approximations help with plotting, but exact forms should be presented when possible.

🧪 Irrational and complex solutions

If the discriminant of the u-equation is positive but not a perfect square, expect irrational solutions for u (and thus for the original variable).
If the discriminant is negative, u will have complex solutions, which may or may not lead to real solutions for the original variable (depends on the substitution).

📐 Finding all roots of polynomials

📐 Factoring higher-degree polynomials

For polynomials like x³ − 8, recognize special forms:

Difference of cubes: a³ − b³ = (a − b)(a² + ab + b²).
Example: x³ − 8 = (x − 2)(x² + 2x + 4).
Set each factor to zero: x − 2 = 0 gives x = 2; x² + 2x + 4 = 0 requires the quadratic formula, yielding x = −1 ± i√3.
This method finds all three roots (one real, two complex), satisfying the fundamental theorem.

📐 Why extracting cube roots is incomplete

Simply taking the cube root (x = ∛8 = 2) finds only the real root, missing the two complex roots that also satisfy the equation.

Solving Quadratic Inequalities

🧭 Overview

🧠 One-sentence thesis

Quadratic inequalities can be solved systematically by finding critical numbers (roots), using a sign chart to determine where the function is positive or negative, and then shading the appropriate regions based on the inequality symbol.

📌 Key points (3–5)

What a quadratic inequality is: a mathematical statement comparing a quadratic expression to zero using inequality symbols (less than, greater than, or their "or equal to" variants).
Critical numbers are the key: the roots of the quadratic function partition the number line into regions where the function is entirely positive or entirely negative.
Sign charts streamline solving: test one value per region to determine the sign, then shade regions that satisfy the inequality without needing a full graph.
Common confusion—strict vs non-strict inequalities: strict inequalities (< or >) use open dots/parentheses; non-strict (≤ or ≥) use closed dots/brackets to include the boundary.
Special cases exist: a quadratic may have no real roots (function always positive or always negative), one repeated root, or infinitely many solutions.

🔍 What quadratic inequalities are

🔍 Definition and solutions

Quadratic inequality: a mathematical statement that relates a quadratic expression as either less than or greater than another.

Examples from the excerpt: x squared minus 2x minus 11 less than or equal to zero; 2x squared minus 7x plus 3 greater than zero; 9 minus x squared greater than zero.
A solution is any real number that makes the inequality true when substituted for the variable.
Solutions can be: infinitely many (an interval or union of intervals), exactly one number, or no solution at all.

🔍 Checking solutions by substitution

Substitute the candidate value into the inequality and simplify.
Example: to check if x = -2 solves x squared minus x minus 6 ≤ 0, compute (-2) squared minus (-2) minus 6 = 4 + 2 - 6 = 0, which satisfies ≤ 0, so -2 is a solution.
If the result does not satisfy the inequality symbol, the value is not a solution.

🎯 Critical numbers and regions

🎯 What critical numbers are

Critical numbers: the values in the domain of a function that separate regions producing positive or negative results.

For quadratic functions, critical numbers are the roots (also called zeros)—where the function equals zero.
Example: f(x) = x squared minus x minus 6 = (x + 2)(x - 3) has roots -2 and 3.
These roots divide the number line into regions; within each region the function does not change sign.

🎯 Geometric interpretation

The graph of a quadratic function is a parabola.
Roots are the x-intercepts (where the parabola crosses the x-axis).
Between roots, the function is either entirely above the x-axis (positive) or entirely below (negative).
Example: if roots are -2 and 3, the function might be positive outside [-2, 3] and negative inside, or vice versa depending on whether the parabola opens upward or downward.

📊 Using sign charts to solve

📊 What a sign chart is

Sign chart: a model of a function using a number line and signs (+ or −) to indicate regions in the domain where the function is positive or negative.

The number line represents the x-axis; critical numbers are marked as boundaries.
Plus signs above a region mean the function is positive there; minus signs mean negative.
Sign charts are useful when a detailed graph is not needed.

📊 Step-by-step solving process

Step 1: Find critical numbers

Rewrite the inequality in standard form (zero on one side).
Set the quadratic expression equal to zero and solve (by factoring, quadratic formula, etc.).
These solutions are the critical numbers.

Step 2: Create the sign chart

Mark the critical numbers on a number line; they partition it into regions.
Choose one test value in each region (any convenient number).
Evaluate the quadratic function at each test value—only the sign (+ or −) matters, not the exact number.
Place + or − signs above each region accordingly.

Step 3: Shade the solution set

If the inequality is "less than zero," shade regions with − signs.
If "greater than zero," shade regions with + signs.
Use open dots (parentheses in interval notation) for strict inequalities (< or >).
Use closed dots (brackets) for non-strict inequalities (≤ or ≥).

📊 Example walkthrough

Solve: negative x squared plus 6x plus 7 ≥ 0.
Factor: -(x + 1)(x - 7) = 0 gives roots x = -1 and x = 7.
Test x = -3 (left region): result is negative.
Test x = 0 (middle region): result is positive.
Test x = 10 (right region): result is negative.
Sign chart: − over (-∞, -1), + over (-1, 7), − over (7, ∞).
The inequality asks for ≥ 0 (positive or zero), so shade the middle region including boundaries: [-1, 7].

🧮 Special cases and techniques

🧮 When roots are irrational

If the quadratic does not factor, use the quadratic formula: x = [negative b plus or minus square root of (b squared minus 4ac)] divided by 2a.
Example: x squared minus 2x minus 11 = 0 gives x = 1 minus 2 times square root of 3 (approximately -2.5) and x = 1 plus 2 times square root of 3 (approximately 4.5).
Proceed with the sign chart using these approximate critical numbers.

🧮 No real roots

If the discriminant (b squared minus 4ac) is negative, there are no real roots—no critical numbers.
The parabola does not cross the x-axis; it is entirely above or entirely below.
Test any single value to determine the sign everywhere.
Example: x squared minus 2x plus 3 > 0 has no real roots; testing x = 0 gives positive result, so the function is positive everywhere. Solution: all real numbers, (-∞, ∞).
If the inequality were x squared minus 2x plus 3 < 0, the answer would be no solution (the function is never negative).

🧮 One repeated root

If the quadratic is a perfect square, there is one repeated root.
Example: 9x squared minus 12x plus 4 = (3x - 2) squared = 0 gives x = 2/3 (one solution).
The function touches the x-axis at exactly one point and is positive (or negative) everywhere else.
For ≤ 0 or ≥ 0, the single root may be the only solution or part of a larger set depending on the parabola's orientation.

🧮 Application to domain of square root functions

The argument of a square root must be non-negative.
To find the domain of f(x) = square root of (x squared minus 4), solve x squared minus 4 ≥ 0.
Roots are x = ±2; testing shows the function is positive outside [-2, 2].
Domain: (-∞, -2] ∪ [2, ∞).

🔄 Comparing solution methods

Method	When to use	Key feature
Graphing the parabola	Visual learners; when graph is already available	Directly see where function is above/below x-axis
Sign chart	Efficient for exams; no graph needed	Only requires sign of test values, not exact computation
Substitution check	Verifying specific values	Confirms whether a particular number is a solution

Don't confuse:

Solving a quadratic equation (find exact x-values where expression = 0) vs solving a quadratic inequality (find intervals where expression satisfies <, >, ≤, or ≥ 0).
Strict inequalities exclude boundaries (open intervals); non-strict include them (closed intervals).

Solving Polynomial and Rational Inequalities

🧭 Overview

🧠 One-sentence thesis

Polynomial and rational inequalities can be solved systematically by finding critical numbers (roots and restrictions), creating a sign chart to test intervals, and identifying where the function satisfies the inequality condition.

📌 Key points (3–5)

Critical numbers differ by type: polynomial inequalities use roots as critical numbers; rational inequalities use both roots (numerator zeros) and restrictions (denominator zeros).
Sign charts reveal solution regions: test values in each interval partitioned by critical numbers to determine where the function is positive or negative.
Dot notation matters: use open dots for strict inequalities (< or >) and closed dots for inclusive inequalities (≤ or ≥); always use open dots for restrictions in rational inequalities.
Common confusion: restrictions vs. roots—restrictions (values that make the denominator zero) are never included in the solution set, even with inclusive inequalities.
Standard form is required: rewrite inequalities with zero on one side before finding critical numbers.

🔢 Polynomial Inequalities

🔢 What defines a polynomial inequality

Polynomial inequality: a mathematical statement that relates a polynomial expression as either less than or greater than another.

The inequality compares a polynomial expression to zero (after rearranging).
Solution involves finding where the polynomial is positive or negative.

🎯 Critical numbers for polynomials

Critical numbers are the roots: values where the polynomial equals zero.
These roots partition the number line into regions.
Example: for x(x + 3)²(x − 4) < 0, the roots are 0, −3, and 4.

📊 Plotting critical numbers

Strict inequalities (< or >): use open dots on the number line.
Inclusive inequalities (≤ or ≥): use closed dots.
The dots indicate whether the critical number itself is included in the solution.

🧮 Creating and Using Sign Charts

🧮 How to build a sign chart

Four-step process:

Obtain zero on one side: rearrange the inequality so all terms are on one side, zero on the other.
Find critical numbers: solve for roots by factoring or other methods.
Create the sign chart: test a value from each interval to determine if the function is positive or negative there.
Answer the question: shade intervals that satisfy the inequality and express in interval notation.

🧪 Testing intervals

Choose any convenient test value within each interval.
Only the sign (+ or −) of the result matters, not the actual value.
Example: if testing x = −5 in (−5)(−5 + 3)²(−5 − 4), calculate (−)(−)²(−) = (+)(−) = (−), so the function is negative in that interval.

✅ Interpreting results

For f(x) < 0: select intervals where the function is negative.
For f(x) > 0: select intervals where the function is positive.
For f(x) ≤ 0 or f(x) ≥ 0: include the critical numbers (closed dots) in the solution.

🔀 Rational Inequalities

🔀 What defines a rational inequality

Rational inequality: a mathematical statement that relates a rational expression as either less than or greater than another.

Involves fractions with polynomials in numerator and denominator.
Requires special attention to domain restrictions.

🚫 Critical numbers for rational functions

Two types of critical numbers:

Type	Definition	Found by	Included in solution?
Roots	Numerator equals zero	Set numerator = 0	Yes (if inequality is inclusive)
Restrictions	Denominator equals zero	Set denominator = 0	Never (not in domain)

Don't confuse: restrictions are critical numbers for partitioning intervals but are never part of the solution set.
Always use open dots for restrictions, regardless of the inequality symbol.

🔧 Standard form requirement

Must have zero on one side before proceeding.
Simplify to a single algebraic fraction.
Example: to solve 7/(x + 3) < 2, rewrite as 7/(x + 3) − 2 < 0, then combine into (−2x + 1)/(x + 3) < 0.

📍 Vertical asymptotes

Restrictions correspond to vertical asymptotes in the graph.
These asymptotes bound regions where the function changes sign.
The function is undefined at restrictions, creating natural boundaries between intervals.

🎓 Worked Example Pattern

🎓 Polynomial example structure

From the excerpt's Example 2 (solving 2x⁴ > 3x³ + 9x²):

Rearrange: 2x⁴ − 3x³ − 9x² > 0
Factor: x²(2x + 3)(x − 3) = 0 gives critical numbers −3/2, 0, and 3
Test intervals with values −2, −1, 1, and 4
Solution: intervals where function is positive: (−∞, −3/2) ∪ (3, ∞)

🎓 Rational example structure

From the excerpt's Example 5 (solving 7/(x + 3) < 2):

Obtain zero on right: 7/(x + 3) − 2 < 0
Combine to single fraction: (−2x + 1)/(x + 3) < 0
Find critical numbers: root at x = 1/2 (numerator), restriction at x = −3 (denominator)
Test intervals and find where negative: (−∞, −3) ∪ (1/2, ∞)

Key distinction: the restriction −3 uses an open dot and is not included, even though it's a boundary of the solution region.

Composition and Inverse Functions

🧭 Overview

🧠 One-sentence thesis

Function composition applies one function to the result of another, and inverse functions reverse each other's effects—a relationship that exists only when a function is one-to-one.

📌 Key points (3–5)

Composition operator (○): notation for applying one function to the output of another; (f ○ g)(x) means substitute g(x) into f(x).
Inverse functions: two functions f and g are inverses if composing them in either order returns x; notation f⁻¹ means "f inverse," not 1/f(x).
One-to-one requirement: a function has an inverse only if it is one-to-one (each range value corresponds to exactly one domain value).
Common confusion: composition is not commutative—(f ○ g)(x) usually does not equal (g ○ f)(x); also, f⁻¹(x) is not the same as 1/f(x).
Geometric symmetry: graphs of inverse functions are mirror images across the line y = x; if (a, b) is on f, then (b, a) is on f⁻¹.

🔗 Composition of functions

🔗 What composition means

Composition of functions: applying a function to the results of another function.

The composition operator (○) indicates this sequential application.
Notation: (f ○ g)(x) = f(g(x)) reads "f composed with g."
Process: first evaluate g at x, then apply f to that result.
Example: if f(x) = x² and g(x) = 2x + 5, then f(g(−1)) means first compute g(−1) = 3, then f(3) = 9.

🔄 Order matters

Composition is not commutative: (f ○ g)(x) and (g ○ f)(x) usually give different results.
Example from the excerpt: with f(x) = x² − x + 3 and g(x) = 2x − 1, we get (f ○ g)(x) = 4x² − 6x + 5 but (g ○ f)(x) = 2x² − 2x + 5.
Don't confuse: the order in which you compose matters; always work from the inside out in the notation f(g(x)).

🔁 Self-composition

A function can be composed with itself: (f ○ f)(x) = f(f(x)).
Example: if f(x) = x² − 2, then (f ○ f)(x) = (x² − 2)² − 2 = x⁴ − 4x² + 2.

🔄 Inverse functions

🔄 What inverse means

Inverse functions: two functions f and g such that (f ○ g)(x) = x and (g ○ f)(x) = x for all x in their respective domains.

Each function "undoes" the effect of the other.
Example: Celsius-to-Fahrenheit and Fahrenheit-to-Celsius conversions are inverses; C(77) = 25 and F(25) = 77.
Notation: if g is the inverse of f, write g = f⁻¹ (read "f inverse").

⚠️ Notation warning

f⁻¹(x) does NOT mean 1/f(x).
The superscript −1 indicates inverse function, not a negative exponent.
This is a common source of confusion; always interpret f⁻¹ as "the function that reverses f."

🪞 Geometric properties

Two key visual features of inverse functions:

Property	Description
Symmetry	Graphs of f and f⁻¹ are mirror images across the line y = x
Point reflection	If (a, b) is on the graph of f, then (b, a) is on the graph of f⁻¹

Example: if (20, 5) is on f, then (5, 20) is on f⁻¹.

🎯 One-to-one functions

🎯 Definition and importance

One-to-one function: a function where each value in the range corresponds to exactly one element in the domain.

A function has an inverse if and only if it is one-to-one.
Not all functions are one-to-one; for example, f(x) = |x| is not (both x = 2 and x = −2 give f(x) = 2).

📏 Horizontal line test

Horizontal line test: if a horizontal line intersects the graph of a function more than once, then it is not one-to-one.

This test determines whether a function has an inverse.
How it works: a horizontal line represents one range value; multiple intersections mean that value corresponds to multiple domain values.
Example: f(x) = x³ passes the test (one-to-one), but f(x) = |x| fails (not one-to-one).
Don't confuse with the vertical line test, which determines whether a graph represents a function at all.

🔒 Restricting domains

If a function is not one-to-one on its full domain, you can sometimes restrict the domain to make it one-to-one.
Example: g(x) = x² + 1 is not one-to-one on all real numbers, but if you restrict to x ≥ 0, it becomes one-to-one and has an inverse.

🔧 Finding inverses algebraically

🔧 Five-step process

The excerpt outlines a systematic method:

Replace f(x) with y: rewrite the function equation using y.
Interchange x and y: swap every x with y and every y with x (reflects the point-swapping property).
Solve for y: isolate y on one side of the equation.
Replace y with f⁻¹(x): the resulting function is the inverse.
Check: verify that (f ○ f⁻¹)(x) = x and (f⁻¹ ○ f)(x) = x.

📐 Linear example

For f(x) = (3/2)x − 5:

After swapping: x = (3/2)y − 5
Solve: y = (2/3)x + 10/3
Result: f⁻¹(x) = (2/3)x + 10/3
Any linear function f(x) = mx + b where m ≠ 0 is one-to-one and has an inverse.

🌀 More complex cases

With restricted domain: for g(x) = x² + 1 where x ≥ 0, the inverse is g⁻¹(x) = √(x − 1).
- Since y ≥ 0 in the original restriction, take only the positive square root.
Rational functions: for f(x) = (2x + 1)/(x − 3), interchange and solve by cross-multiplying, then factor out y as a common factor to isolate it.
- Result: f⁻¹(x) = (3x + 1)/(x − 2).

✅ Verification

Always check both compositions: (f ○ f⁻¹)(x) = x and (f⁻¹ ○ f)(x) = x.
If both simplify to x, the functions are confirmed as inverses.
Example: verifying f(x) = (1/2)x − 5 and g(x) = 2x + 10 requires showing f(g(x)) = x and g(f(x)) = x by substitution and simplification.

Exponential Functions and Their Graphs

🧭 Overview

🧠 One-sentence thesis

Exponential functions—where the variable appears in the exponent—grow or decay at rates determined by their base, and they model real-world phenomena like compound interest with domains spanning all real numbers and ranges restricted to positive values.

📌 Key points (3–5)

What exponential functions are: functions of the form f(x) = b^x where b > 0 and b ≠ 1, with the variable in the exponent rather than the base.
Domain and range: domain is all real numbers; range is all positive numbers (0, ∞), with a horizontal asymptote at y = 0.
Growth vs decay behavior: bases greater than 1 produce growth (increasing graphs); bases between 0 and 1 produce decay (decreasing graphs).
Common confusion—base restrictions: the base b = 1 is excluded because 1^x = 1 for all x (constant, not exponential); bases must be positive to keep outputs real.
Real applications: compound interest formulas use exponential functions, with continuous compounding involving the natural base e ≈ 2.71828.

📐 Definition and basic structure

📐 What makes a function exponential

Exponential function: any function with definition of the form f(x) = b^x where b > 0 and b ≠ 1.

The key feature is that the variable is in the exponent, not the base.
Contrast with power functions like x^2 or y^(-3), where the base is variable and the exponent is constant.
Example: f(x) = 2^x is exponential; the input x determines the power to which 2 is raised.

🌐 Domain and range characteristics

Domain: all real numbers (−∞, ∞).
- Rational exponents have always been defined; irrational exponents (like 2^√7) can be approximated to any accuracy using rational bounds.
- The excerpt shows 2.64 < √7 < 2.65, so 2^2.64 < 2^√7 < 2^2.65, yielding an approximation.
Range: all positive numbers (0, ∞).
- As x approaches negative infinity, b^x approaches zero but never reaches it.
- Example: f(−5) = 2^(−5) = 1/32 ≈ 0.03125; f(−15) ≈ 0.00003052—very small but never zero.
Horizontal asymptote: y = 0 (the x-axis) is the lower bound.

🎯 The y-intercept rule

Every exponential function f(x) = b^x passes through the point (0, 1).
Why: f(0) = b^0 = 1 for any base b.
This is true regardless of the base value.

📈 Growth and decay patterns

📈 When the base is greater than 1 (b > 1)

The graph increases (grows) as you read from left to right.
Larger bases grow faster.
Example comparison from the excerpt:
- y = 2^x, y = 3^x, and y = 10^x all pass through (0, 1).
- For the same x > 0, the function with base 10 produces the largest output.
The excerpt calls base 10 the common base because it appears frequently in scientific notation.

📉 When the base is a fraction (0 < b < 1)

The graph decreases (decays) as you read from left to right.
Example: f(x) = (1/2)^x.
- At x = −2: (1/2)^(−2) = 2^2 = 4.
- At x = 2: (1/2)^2 = 1/4.
Relationship to negative exponents: (1/2)^x = 2^(−x).
- A fractional-base exponential function can be rewritten with a negative exponent and a base greater than 1.
- Geometrically, g(x) = 2^(−x) is a reflection of f(x) = 2^x about the y-axis.

🔄 One-to-one property

All exponential functions pass the horizontal line test.
This means they are one-to-one: each output corresponds to exactly one input.
This property is important for defining inverse functions (logarithms, covered later).

🌿 The natural base e

🌿 What e represents

e ≈ 2.71828: an irrational constant called the natural base or Euler's constant.
It arises naturally in growth and decay models.
The excerpt notes it is named after Leonhard Euler (pronounced "Oiler").

🌿 The natural exponential function

Natural exponential function: f(x) = e^x.

Most scientific calculators have a dedicated e^x button.
The graph of y = e^x is similar to y = 3^x.
Domain: (−∞, ∞); Range: (0, ∞); y-intercept: (0, 1); asymptote: y = 0.
Example values (rounded):
- f(−2) = e^(−2) ≈ 0.14
- f(0) = 1
- f(2) = e^2 ≈ 7.39

🔗 Connection to continuous compounding

The number e emerges when compounding interest more and more frequently.
The excerpt shows a table: as n (compounding frequency) increases in (1 + 1/n)^n, the result approaches e.
- Yearly (n = 1): 2
- Monthly (n = 12): ≈ 2.61304
- Daily (n = 365): ≈ 2.71457
- Hourly (n = 8760): ≈ 2.71813
As n → ∞ (compounding every instant), the limit is e.

🎨 Graphing with transformations

🎨 Vertical and horizontal shifts

Vertical shift: f(x) = b^x + k shifts the graph up (k > 0) or down (k < 0).
- The horizontal asymptote moves from y = 0 to y = k.
- Example: f(x) = 10^x + 5 has asymptote y = 5 and range (5, ∞).
Horizontal shift: f(x) = b^(x − h) shifts right (h > 0) or left (h < 0).
- Example: f(x) = 2^(x − 3) shifts the basic graph 3 units right.

🎨 Reflections

Reflection about the x-axis: f(x) = −b^x flips the graph upside down.
- Range becomes (−∞, 0); asymptote remains y = 0.
- Example: g(x) = −2^(x−3) is reflected about the x-axis and shifted right 3 units.
Reflection about the y-axis: f(x) = b^(−x) is equivalent to using a reciprocal base.
- Example: 2^(−x) = (1/2)^x.

🎨 Finding the y-intercept after transformations

Always set x = 0 and evaluate.
Example from the excerpt: g(x) = −2^(x−3).
- g(0) = −2^(0−3) = −2^(−3) = −1/8.
- Y-intercept: (0, −1/8).

💰 Compound interest applications

💰 Periodic compounding formula

Compound interest formula: A(t) = P(1 + r/n)^(nt).

P: principal (initial investment).
r: annual interest rate (as a decimal).
n: number of times interest is compounded per year.
t: time in years.
A(t): amount accumulated after t years.

Example from the excerpt: $500 invested at 4.5% annual interest compounded monthly for 6 years.

P = 500, r = 0.045, n = 12, t = 6.
A(6) = 500(1 + 0.045/12)^(12·6) = 500(1.00375)^72 ≈ $654.65.

💰 Continuous compounding formula

Continuously compounding interest formula: A(t) = Pe^(rt).

Used when interest is compounded every instant (as n → ∞).
P: principal.
r: annual interest rate.
t: time in years.

Example from the excerpt: Same $500 at 4.5% for 6 years, compounded continuously.

A(6) = 500e^(0.045·6) = 500e^0.27 ≈ $654.98.

💰 Comparing compounding methods

The excerpt compares the two examples: monthly compounding yields $654.65; continuous compounding yields $654.98.
Key insight: "Compounding continuously may not be as beneficial as it sounds"—the difference is small.
The interest rate r has a much greater impact than the compounding frequency.

🔑 Key properties summary

Property	When b > 1	When 0 < b < 1
Behavior	Increasing (growth)	Decreasing (decay)
Domain	(−∞, ∞)	(−∞, ∞)
Range	(0, ∞)	(0, ∞)
y-intercept	(0, 1)	(0, 1)
Asymptote	y = 0	y = 0
One-to-one?	Yes	Yes

🔑 Why b = 1 is excluded

If b = 1, then f(x) = 1^x = 1 for all x.
This is a constant function, not an exponential function.
The excerpt explicitly states b ≠ 1 in the definition.

🔑 Why b must be positive

Negative bases can produce non-real outputs for certain exponents.
Example: (−4)^(1/2) = √(−4) is not a real number.
Restricting b > 0 ensures the range consists of real positive numbers.

Logarithmic Functions and Their Graphs

🧭 Overview

🧠 One-sentence thesis

Logarithmic functions are defined as the inverses of exponential functions, allowing us to solve for exponents and graph curves that mirror exponential behavior across the line y = x.

📌 Key points (3–5)

What a logarithm is: the exponent to which a base must be raised to obtain a specific value; y = log base b of x means b raised to y equals x.
Two special logarithms: the common logarithm (base 10, written as log x) and the natural logarithm (base e, written as ln x) are widely used in applications.
Domain and range: logarithmic functions accept only positive real numbers as inputs (domain: (0, ∞)) but produce all real numbers as outputs (range: (−∞, ∞)).
Common confusion: the argument of a logarithm can never be negative or zero—there is no power of a positive base that yields a negative number or zero.
Graphing behavior: when the base b > 1, the graph rises from left to right with a vertical asymptote at x = 0; when the base is a fraction (0 < b < 1), the graph falls from left to right.

🔄 Definition and inverse relationship

🔄 Logarithm as inverse of exponential

The logarithm base b is defined as the inverse of the exponential function with base b.

If f(x) = 2 raised to x is one-to-one (passes the horizontal line test), then its inverse exists.
Reflecting the exponential graph about the line y = x produces the logarithmic graph.
The inverse function is written as f inverse (x) = log base 2 of x.

🔁 Converting between forms

y = log base b of x if and only if x = b raised to y.

This equivalence allows you to rewrite logarithmic statements as exponential statements and vice versa.
Example: log base 2 of 16 = 4 is equivalent to 2 raised to 4 = 16.
Example: log base 5 of (1/25) = −2 is equivalent to 5 raised to −2 = 1/25.

🎯 What the logarithm represents

The logarithm is actually the exponent y to which the base b is raised to obtain the argument x.
Example: log base 5 of 125 = 3 because 5 raised to 3 = 125.
Example: log base 2 of (1/8) = −3 because 2 raised to −3 = 1/8.
Don't confuse: the logarithm is not the result of raising the base to a power; it is the power itself.

🔢 Common and natural logarithms

🔟 Common logarithm (base 10)

The common logarithm is the logarithm base 10, denoted log x (without an explicit base).

When a logarithm is written without a base, it is assumed to be base 10.
Example: log 1000 = 3 because 10 raised to 3 = 1000.
Example: log 0.01 = −2 because 10 raised to −2 = 1/100 = 0.01.
Calculators have a LOG button for evaluating common logarithms of numbers like 75, which yield non-integer results (log 75 ≈ 1.875).

🌿 Natural logarithm (base e)

The natural logarithm is the logarithm base e, denoted ln x.

The natural logarithm is widely used in mathematics and science.
Example: ln e = 1 because e raised to 1 = e.
Example: ln (1/e raised to 4) = −4 because e raised to −4 = 1/e raised to 4.
Calculators have an LN button for evaluating natural logarithms (ln 75 ≈ 4.317).

🧮 Using calculators

For results that are not obvious, use the LOG or LN button.
Example: to find x when log x = 3.2, convert to exponential form: x = 10 raised to 3.2 ≈ 1584.893.
Example: to find x when ln x = −4, convert to exponential form: x = e raised to −4 ≈ 0.018.

📊 Graphing logarithmic functions

📈 Basic shape when base b > 1

The graph of f(x) = log base b of x has the following features:
- Domain: (0, ∞) — only positive real numbers
- Range: (−∞, ∞) — all real numbers
- Vertical asymptote: x = 0 (the y-axis)
- x-intercept: (1, 0)
- The point (b, 1) is always on the graph because log base b of b = 1.

📉 Basic shape when 0 < base < 1

When the base is a fraction (e.g., f(x) = log base (1/2) of x), the graph falls from left to right.
The domain, range, asymptote, and x-intercept remain the same as when b > 1.
The point (b, −1) is on the graph because log base (1/b) of b = −1.

🔀 Transformations

Horizontal shifts: f(x) = log base b of (x + h) shifts the graph h units left (if h > 0) or right (if h < 0).
- The vertical asymptote also shifts to x = −h.
- Example: f(x) = log base 3 of (x + 4) has asymptote at x = −4 and domain (−4, ∞).
Vertical shifts: f(x) = log base b of x + k shifts the graph k units up (if k > 0) or down (if k < 0).
Reflections: f(x) = −log base b of x reflects the graph about the x-axis.
- Example: f(x) = −log (x − 2) is reflected and shifted right 2 units, with domain (2, ∞).
Don't confuse: the asymptote moves with horizontal shifts but not with vertical shifts or reflections.

🎨 Sketching strategy

Identify the basic graph (log base b of x).
Apply transformations in order: reflections, horizontal shifts, vertical shifts.
Draw the vertical asymptote with a dashed line—it defines the boundary of the domain.
Mark key points such as the x-intercept and any known points like (b, 1).

⚠️ Domain restrictions and common errors

⚠️ Why negative arguments are undefined

Negative numbers and zero are not in the domain of the logarithm.
There is no power of a positive base that results in a negative number or zero.
Example: log base 2 of (−4) is undefined because 2 raised to any real number cannot equal −4.
Example: log base 2 of 0 is undefined because 2 raised to any real number cannot equal 0.

⚠️ The result of a logarithm can be negative

Don't confuse the argument (input) with the result (output).
The argument must be positive, but the result can be negative, zero, or positive.
Example: log base 2 of (1/8) = −3 is valid because the argument 1/8 is positive, even though the result is negative.

⚠️ Base restrictions

The base b must satisfy b > 0 and b ≠ 1.
b = 1 is excluded because 1 raised to any power is always 1, so the function would not be one-to-one and would have no inverse.

🌍 Applications

🌍 Richter scale (earthquakes)

The magnitude M of an earthquake is given by M = log (I / I₀), where I is the measured intensity and I₀ is the minimum intensity.
Example: if an earthquake intensity is 100 times the minimum, then M = log 100 = 2.
Example: if intensity is 3 million times the minimum, then M = log 3,000,000 ≈ 6.5.

🌍 pH scale (acidity)

In chemistry, pH = −log (H⁺), where H⁺ is the hydrogen ion concentration.
Example: pure water has H⁺ = 0.0000001, so pH = −log (0.0000001) = 7.
Example: lemon juice has H⁺ = 0.01, so pH = −log (0.01) = 2.

Properties of the Logarithm

🧭 Overview

🧠 One-sentence thesis

The logarithm has inverse properties that simplify expressions and three core expansion rules—product, quotient, and power—that allow us to rewrite complex logarithmic expressions as sums, differences, and coefficients or combine multiple logarithms into a single logarithm.

📌 Key points (3–5)

Inverse properties: log base b of b to the x equals x, and b to the log base b of x equals x (when x > 0), allowing direct simplification.
Product property: the logarithm of a product equals the sum of the logarithms of the factors.
Quotient property: the logarithm of a quotient equals the difference of the logarithms (numerator minus denominator).
Power property: the logarithm of a quantity raised to a power equals that power times the logarithm of the quantity.
Common confusion: log of (x + y) does NOT equal log x + log y; there is no rule for expanding the logarithm of a sum or difference.

🔄 Inverse properties and basic values

🔄 Definition and inverse relationship

The base-b logarithm is defined as: y = log base b of x if and only if x = b to the y (where b > 0 and b ≠ 1).

Because logarithms and exponentials are inverses, applying one after the other returns the original input.
The excerpt derives two inverse properties from this definition.

🔢 Special logarithm values

From the definition, the excerpt establishes these basic results:

Expression	Value	Reason
log base b of 1	0	b to the 0 equals 1
log base b of b	1	b to the 1 equals b
log base b of (1/b)	−1	b to the −1 equals 1/b
log base (1/b) of b	−1	(1/b) to the −1 equals b

Example: log 1 = 0 (common logarithm, base 10); ln e = 1 (natural logarithm, base e); log base 5 of (1/5) = −1.
Don't confuse: the base matters—log base (1/4) of 4 = −1 because (1/4) to the −1 equals 4.

⚡ The two inverse properties

The excerpt highlights these as the inverse properties of the logarithm:

log base b of (b to the x) = x for all x.
b to the (log base b of x) = x when x > 0.

The restriction x > 0 comes from the domain of the logarithm (logarithms are only defined for positive arguments).
Example: log base 5 of 625 = log base 5 of (5 to the 4) = 4; 5 to the (log base 5 of 3) = 3; e to the (ln 5) = 5.

📦 Product and quotient properties

📦 Product property

Product property of logarithms: log base b of (xy) = log base b of x + log base b of y.

In words: the logarithm of a product is equal to the sum of the logarithm of the factors.
The excerpt derives this by setting u = log base b of x and v = log base b of y, converting to exponential form (x = b to the u, y = b to the v), then substituting into log base b of (xy) and using exponent rules (b to the u times b to the v = b to the (u + v)).
Example: log base 2 of (8x) = log base 2 of 8 + log base 2 of x = 3 + log base 2 of x.

➗ Quotient property

Quotient property of logarithms: log base b of (x/y) = log base b of x − log base b of y.

In words: the logarithm of a quotient is equal to the difference of the logarithm of the numerator and the logarithm of the denominator.
Derived similarly using exponent rules (b to the u divided by b to the v = b to the (u − v)).
Example: log of (x/10) = log x − log 10 = log x − 1.
Caution: When applying the quotient property, distribute the negative sign carefully; log base 2 of ((x+1) squared / (5y)) = log base 2 of (x+1) squared − (log base 2 of 5 + log base 2 of y).

🔺 Power property

Power property of logarithms: log base b of (x to the n) = n times log base b of x.

In words: the logarithm of a quantity raised to a power is equal to that power times the logarithm of the quantity.
The excerpt derives this by starting with log base b of x = u, converting to exponential form, raising both sides to the nth power, then converting back.
Example: log base 2 of (x to the 4) = 4 log base 2 of x; log base 5 of (square root of x) = log base 5 of (x to the 1/2) = (1/2) log base 5 of x.

🌳 Applying to natural logarithms

The natural logarithm (ln x = log base e of x) is just a logarithm with base e, so all three properties apply.
Example: ln(2x cubed) = ln 2 + ln(x cubed) = ln 2 + 3 ln x.

🔍 Expanding logarithmic expressions

🔍 What "expand completely" means

A logarithmic expression is completely expanded when the properties of the logarithm can no further be applied.
The goal is to rewrite a single logarithm involving products, quotients, and powers as a combination of sums, differences, and coefficients.

🧮 Multi-step expansion

Example from the excerpt: expand log of (cube root of (10xy squared)).

Rewrite the cube root as a rational exponent: (10xy squared) to the (1/3).
Apply the power property: (1/3) log(10xy squared).
Apply the product property: (1/3)(log 10 + log x + log(y squared)).
Apply the power property again: (1/3)(1 + log x + 2 log y).
Distribute: 1/3 + (1/3) log x + (2/3) log y.

⚠️ Critical warning: no rule for sums or differences

The excerpt emphasizes this caution multiple times:

log(x + y) ≠ log x + log y
log(x / y) ≠ (log x) / (log y)
log(xy) ≠ (log x)(log y)

There is no property that allows you to expand the logarithm of a sum or difference.

Example: log base 2 of ((x+1) squared) cannot be split into log base 2 of x + log base 2 of 1; you can only apply the power property to get 2 log base 2 of (x+1).

🔗 Combining logarithms (condensing)

🔗 Writing as a single logarithm

The reverse process: combine an expression involving multiple logarithms into a single logarithm with coefficient 1.
The excerpt notes this is "an essential skill" and "one of the first steps when solving logarithmic equations."

🛠️ Step-by-step condensing

Example from the excerpt: write 3 log base 3 of x − log base 3 of y + 2 log base 3 of 5 as a single logarithm.

Use the power property to rewrite coefficients as exponents: log base 3 of (x cubed) − log base 3 of y + log base 3 of (5 squared).
Apply the quotient property to the first two terms: log base 3 of (x cubed / y) + log base 3 of 25.
Apply the product property: log base 3 of ((x cubed / y) times 25) = log base 3 of (25x cubed / y).

🧩 Working with natural logarithms

Example: (1/2) ln x − 3 ln y − ln z.

Rewrite as ln(x to the 1/2) − ln(y cubed) − ln z.
Apply quotient property twice (left to right): ln((x to the 1/2) / (y cubed)) − ln z = ln((x to the 1/2) / (y cubed times z)).
Simplify: ln(square root of x / (y cubed times z)).

📋 Summary of all properties

The excerpt concludes with a complete list when b > 0 and b ≠ 1:

Property	Formula
Log of 1	log base b of 1 = 0
Log of base	log base b of b = 1
Log of reciprocal base	log base (1/b) of b = −1 or log base b of (1/b) = −1
Inverse property 1	log base b of (b to the x) = x
Inverse property 2	b to the (log base b of x) = x (x > 0)
Product property	log base b of (xy) = log base b of x + log base b of y
Quotient property	log base b of (x/y) = log base b of x − log base b of y
Power property	log base b of (x to the n) = n log base b of x

All properties apply to the natural logarithm since ln x = log base e of x.

Solving Exponential and Logarithmic Equations

🧭 Overview

🧠 One-sentence thesis

Exponential and logarithmic equations can be solved by exploiting the one-to-one property (when bases match) or by applying logarithms/definitions (when bases differ), but logarithmic solutions require careful checking for extraneous answers.

📌 Key points (3–5)

Two methods for exponentials: use the one-to-one property when both sides share a base; otherwise isolate the exponential and apply a logarithm.
Two methods for logarithms: use the one-to-one property when a single log with the same base appears on each side; otherwise combine into one log and apply the definition.
Change of base formula: allows approximation of any logarithm on a calculator by dividing log of the argument by log of the base.
Common confusion: logarithmic equations often produce extraneous solutions because the domain restricts arguments to positive values—always check your answer.
Why it matters: these techniques solve real-world problems (population growth, pH, sound intensity) and enable finding inverses of exponential/logarithmic functions.

🔑 One-to-one properties

🔑 For exponential functions

One-to-one property of exponential functions: Given b > 0 and b ≠ 1, b^x = b^y if and only if x = y.

Because exponential functions are one-to-one, equal outputs guarantee equal inputs.
When to use: when you can rewrite both sides of an equation with the same base.
Example: To solve 3^(2x−1) = 27, rewrite 27 as 3³, giving 3^(2x−1) = 3³, so 2x − 1 = 3, hence x = 2.
Don't confuse: this only works when the bases match; if bases differ, you must use logarithms instead.

🔑 For logarithmic functions

One-to-one property of logarithms: Given b > 0 and b ≠ 1 where x, y > 0, log_b(x) = log_b(y) if and only if x = y.

Logarithms are also one-to-one, so equal logs (same base) mean equal arguments.
When to use: when you can isolate a single logarithm with the same base on each side.
Example: log₂(2x − 5) = log₂(x − 2) implies 2x − 5 = x − 2, so x = 3.
Critical warning: the arguments must be positive; always verify your solution is in the domain.

🧮 Solving exponential equations

🧮 Method 1: Matching bases

Step 1: Express both sides as powers of the same base.
Step 2: Set exponents equal using the one-to-one property.
Step 3: Solve the resulting equation.
Example: 16^(1−3x) = 2 becomes (2⁴)^(1−3x) = 2¹, so 4(1 − 3x) = 1, giving x = 1/4.

🧮 Method 2: Applying logarithms

When to use: when you cannot easily match the bases.
Step 1: Isolate the exponential expression.
Step 2: Take the logarithm of both sides (common log or natural log).
Step 3: Use the power rule (log(b^x) = x·log(b)) to bring the exponent down.
Step 4: Solve for the variable.
Example: For 5^(2x−1) + 2 = 9, isolate to get 5^(2x−1) = 7, then log(5^(2x−1)) = log(7), so (2x − 1)·log(5) = log(7), yielding x = (log(5) + log(7))/(2·log(5)).
Calculator tip: use parentheses carefully when entering fractions like (log(5) + log(7))/(2·log(5)).

🧮 Natural logarithm shortcut

When the base is e, apply the natural logarithm (ln) to both sides.
Recall ln(e) = 1, so ln(e^(5x+3)) simplifies to (5x + 3)·1 = 5x + 3.
Example: e^(5x+3) = 1 becomes ln(e^(5x+3)) = ln(1), so 5x + 3 = 0, giving x = −3/5.

🔄 Change of base formula

🔄 The formula

Change of base formula: log_a(x) = log_b(x) / log_b(a), where you can choose any base b.

Allows you to compute logarithms of any base using only common log (base 10) or natural log (base e).
Why it works: derived by rewriting y = log_a(x) in exponential form (x = a^y), then applying log_b to both sides and solving for y.

🔄 Using a calculator

Most calculators only have LOG (base 10) and LN (base e) buttons.
To approximate log₃(10), compute log(10)/log(3) ≈ 2.0959 or ln(10)/ln(3) ≈ 2.0959.
The result is independent of whether you use common or natural log.
Example: log₇(120) = log(120)/log(7) ≈ 2.46.

📐 Solving logarithmic equations

📐 Method 1: Matching logarithms

When to use: when you can obtain a single log with the same base on each side.
Use log properties (product, quotient, power rules) to combine multiple logs.
Then apply the one-to-one property and set arguments equal.
Example: log₂(2x − 5) − log₂(x − 2) = 0 becomes log₂(2x − 5) = log₂(x − 2), so 2x − 5 = x − 2, giving x = 3.

📐 Method 2: Using the definition

When to use: when you cannot match logarithms on both sides.
Step 1: Combine all logs into a single logarithm with coefficient 1.
Step 2: Apply the definition: log_b(x) = y means x = b^y.
Step 3: Solve the resulting equation.
Step 4: Check for extraneous solutions (required step).
Example: log₃(2x − 5) = 2 means 2x − 5 = 3² = 9, so x = 7.

📐 Multi-step example

For log₂(x − 2) + log₂(x − 3) = 1:

Step 1: Use product rule: log₂[(x − 2)(x − 3)] = 1.
Step 2: Apply definition: (x − 2)(x − 3) = 2¹ = 2.
Step 3: Expand and solve: x² − 5x + 6 = 2, so x² − 5x + 4 = 0, giving (x − 4)(x − 1) = 0, hence x = 4 or x = 1.
Step 4: Check x = 4: log₂(2) + log₂(1) = 1 + 0 = 1 ✓. Check x = 1: log₂(−1) is undefined ✗.
Only x = 4 is valid; x = 1 is extraneous.

⚠️ Domain restrictions and extraneous solutions

Logarithms are only defined for positive arguments: log(x − 2) requires x > 2.
Solving algebraically may produce values outside this domain.
Don't skip the check: substitute each solution back into the original equation to verify all logarithmic expressions are defined.
Example: log(3x − 4) = log(x − 2) gives x = 1, but log(1 − 2) = log(−1) is undefined, so there is no solution.

🔁 Finding inverses

🔁 Inverse of logarithmic functions

Start with y = log_b(expression in x).
Swap x and y, then apply the definition to solve for y.
Example: For f(x) = log₂(3x − 4), swap to get x = log₂(3y − 4), so 3y − 4 = 2^x, giving y = (2^x + 4)/3, hence f⁻¹(x) = (2^x + 4)/3.

🔁 Inverse of exponential functions

Start with y = b^(expression in x).
Swap x and y, then apply the appropriate logarithm.
Example: For g(x) = e^(3x), swap to get x = e^(3y), so ln(x) = 3y, giving y = ln(x)/3, hence g⁻¹(x) = ln(x)/3.

📊 Summary table

Equation type	When bases/logs match	When they don't match
Exponential b^x = b^y	Set exponents equal: x = y	Apply log to both sides, use power rule
Logarithmic log_b(x) = log_b(y)	Set arguments equal: x = y	Combine into one log, apply definition
Key property	One-to-one property	Power rule / definition
Special care	Verify solution works	Always check domain for logs

Applications

🧭 Overview

🧠 One-sentence thesis

Exponential and logarithmic functions model real-world phenomena such as compound interest, population growth, and radioactive decay, allowing us to calculate future values and time intervals using specific formulas.

📌 Key points (3–5)

Compound vs. continuous interest: Compound interest uses A(t) = P(1 + r/n)^(nt) when compounded n times per year; continuous interest uses A(t) = Pe^(rt).
Doubling time is independent of principal: The time it takes an investment to double depends only on the interest rate and compounding frequency, not the initial amount.
Exponential growth/decay model: P(t) = P₀e^(kt) models populations, bacteria, and radioactive substances, where k > 0 indicates growth and k < 0 indicates decay.
Half-life is independent of initial amount: The time for a substance to decay to half its quantity depends only on the decay rate, not the starting amount.
Common confusion: Don't confuse the number of compoundings per year (n) with continuous compounding; "continuous" is a keyword that tells you to use the e-based formula instead.

💰 Interest formulas

💰 Compound interest formula

Compound interest formula: A(t) = P(1 + r/n)^(nt), where P is the initial principal, r is the annual interest rate, n is the number of times interest is compounded per year, and t is time in years.

This formula applies when interest is added to the principal at regular intervals (monthly, quarterly, semi-annually, etc.).
The value n determines compounding frequency: n = 12 for monthly, n = 4 for quarterly, n = 2 for semi-annually, n = 365 for daily.
Example: Susan invests $500 at 4.5% annual interest compounded monthly (n = 12). After 3 years, the amount is A(3) = 500(1 + 0.045/12)^(12·3) ≈ $572.12.

💰 Continuous compounding formula

Continuous interest formula: A(t) = Pe^(rt), where P is the initial principal, r is the annual interest rate, and t is time in years.

Use this formula when the problem states interest is compounded "continuously."
The base e (Euler's number) represents instantaneous compounding at every moment.
Example: Mary invests $200 at 5.75% annual interest compounded continuously. To find when it reaches $350, solve 350 = 200e^(0.0575t), which gives t ≈ 9.73 years.

💰 Solving for time in interest problems

To find how long an investment takes to reach a target amount:

Set A(t) equal to the target amount
Isolate the exponential expression
Take the logarithm of both sides (common log for base-10, natural log for base-e)
Apply the power rule for logarithms
Solve for t

Example: To find when $750 is reached from $500 at 4.5% compounded monthly:

750 = 500(1.00375)^(12t)
1.5 = (1.00375)^(12t)
log(1.5) = 12t · log(1.00375)
t = log(1.5) / (12 · log(1.00375)) ≈ 9 years

⏱️ Doubling time and half-life

⏱️ Doubling time

Doubling time: The period of time it takes a quantity to double.

Doubling time is independent of the initial investment amount.
Whether you invest $1,000 or $1,000,000, the time to double at the same interest rate is identical.
Example: Mario invests $1,000 at 6.3% compounded semi-annually. To find doubling time, solve 2,000 = 1,000(1.0315)^(2t), which simplifies to 2 = (1.0315)^(2t). The result is t ≈ 11.17 years. Any other principal amount would yield the same doubling time.

⏱️ Half-life

Half-life: The period of time it takes a quantity to decay to one-half of the initial amount.

Half-life is independent of the initial amount of substance.
Used primarily for radioactive decay and other decreasing quantities.
Example: Caesium-137 has a half-life of 30 years. Whether you start with 50 mg or 5 mg, it takes 30 years to decay to half the original amount.

Don't confuse: Doubling time applies to growth (positive rates), while half-life applies to decay (negative rates). Both are independent of initial quantity.

📈 Exponential growth and decay model

📈 The general formula

Exponential growth/decay formula: P(t) = P₀e^(kt), where P₀ is the initial amount, k is the growth/decay rate, and t is time.

If k > 0, the model describes exponential growth (populations, bacteria, investments).
If k < 0, the model describes exponential decay (radioactive substances, depreciation).
This formula is similar to continuous compound interest but applies to any exponentially changing quantity.

📈 Determining the growth rate k

When k is not given directly, use additional information to find it:

Step 1: Use known values to set up an equation.

Example: E. coli bacteria double in 20 minutes. If P₀ = 1,000 cells, then P(20) = 2,000 cells.
Equation: 2,000 = 1,000e^(k·20)

Step 2: Solve for k.

2 = e^(20k)
ln(2) = 20k
k = ln(2)/20 ≈ 0.0347 (about 3.5% growth per minute)

Step 3: Write the model using the exact value of k.

P(t) = 1,000e^((ln(2)/20)t)
Using exact values avoids round-off error in final calculations.

Step 4: Use the model to answer questions.

Example: How many cells after 2 hours (120 minutes)? P(120) = 1,000e^((ln(2)/20)·120) = 1,000 · 2^6 = 64,000 cells.

📈 Population growth applications

Population growth is modeled when conditions are optimal and growth is unrestricted.
Example: A town has 93,000 people with 2.6% annual growth. Model: P(t) = 93,000e^(0.026t).
- In 7 years: P(7) = 93,000e^(0.026·7) ≈ 111,564 people.
- To reach 120,000: solve 120,000 = 93,000e^(0.026t), giving t ≈ 9.8 years.

☢️ Radioactive decay applications

☢️ Using half-life to find decay rate

To model radioactive decay when given half-life:

Step 1: Use half-life to determine k.

Example: Caesium-137 has a half-life of 30 years. Starting with P₀ = 50 mg, after 30 years P(30) = 25 mg.
25 = 50e^(k·30)
0.5 = e^(30k)
ln(0.5) = 30k
k = -ln(2)/30 (negative because it's decay)

Step 2: Write the decay model.

P(t) = 50e^((-ln(2)/30)t)

Step 3: Use the model to find when a specific amount remains.

Example: When does it decay to 10 mg?
10 = 50e^((-ln(2)/30)t)
Solving gives t ≈ 69.66 years.

☢️ Radiocarbon dating

Carbon-14 dating estimates the age of artifacts based on remaining carbon-14.
When an organism dies, carbon-14 stops being absorbed and begins to decay.
Carbon-14 has a half-life of 5,730 years.

Example: A bone tool contains 25% of the original carbon-14.

Model: P(t) = P₀e^((-ln(2)/5730)t)
Set P(t) = 0.25P₀
0.25 = e^((-ln(2)/5730)t)
ln(0.25) = (-ln(2)/5730)t
t = -5730·ln(0.25)/ln(2) ≈ 11,460 years old

Don't confuse: The initial amount P₀ cancels out when solving for time, confirming that half-life (and dating calculations) are independent of the starting quantity.

🔑 Key problem-solving steps

🔑 Identifying which formula to use

Situation	Formula	Key indicators
Compound interest	A(t) = P(1 + r/n)^(nt)	"compounded monthly/quarterly/semi-annually/daily"
Continuous interest	A(t) = Pe^(rt)	"compounded continuously"
Growth/decay	P(t) = P₀e^(kt)	"grows/decays exponentially," population, bacteria, radioactive substances

🔑 General solution strategy

Construct the model: Identify P₀ (or P), r (or k), n (if applicable), and the appropriate formula.
Find unknown rates if needed: Use given information (doubling time, half-life, or two data points) to solve for k or verify r.
Set up the equation: Substitute known values and the target amount.
Isolate the exponential: Divide both sides to get the exponential expression alone.
Apply logarithms: Take ln (for base e) or log (for base 10) of both sides.
Solve for the variable: Use logarithm properties and algebra to isolate t or the unknown.
Calculate and interpret: Use a calculator for numerical answers and state the result with appropriate units.

Distance, Midpoint, and the Parabola

🧭 Overview

🧠 One-sentence thesis

This section teaches how to calculate distances and midpoints between coordinate points and how to graph parabolas by transforming their equations into standard form through completing the square.

📌 Key points (3–5)

Distance and midpoint formulas: tools to find the length between two points and the point halfway between them on a coordinate plane.
Parabolas as conic sections: curves formed by slicing a cone with a plane, defined as points equidistant from a focus and a directrix.
Standard vs general form: parabolas can be written as y = ax² + bx + c (general) or y = a(x − h)² + k (standard), where standard form reveals the vertex directly.
Completing the square: the main technique to convert general form into standard form by adding and subtracting the same value.
Common confusion: parabolas can open in four directions (up, down, left, right); those opening left or right are quadratic in y, not x, and do not represent functions.

📏 Distance and Midpoint Formulas

📏 Distance formula

Given two points (x₁, y₁) and (x₂, y₂), the distance d between them is given by d = √[(x₂ − x₁)² + (y₂ − y₁)²].

This formula calculates the straight-line distance between any two points on a rectangular coordinate plane.
It comes from the Pythagorean theorem applied to the horizontal and vertical differences.
Example: For points (−2, −5) and (−4, −3), the distance is √[(−4 − (−2))² + (−3 − (−5))²] = √[4 + 4] = √8 = 2√2 units.

📍 Midpoint formula

Given two points (x₁, y₁) and (x₂, y₂), the midpoint is the ordered pair ((x₁ + x₂)/2, (y₁ + y₂)/2).

The midpoint is the point that bisects (cuts in half) the line segment between two points.
It is found by averaging the x-values and averaging the y-values separately.
Example: For points (−2, −5) and (−4, −3), the midpoint is ((−2 + (−4))/2, (−5 + (−3))/2) = (−3, −4).

🔧 Application to circles

If two points define a circle's diameter, the distance formula gives the diameter length.
The radius is half the diameter.
The area can then be calculated using A = πr².
Example: If the diameter endpoints are (−1, 2) and (1, −2), the diameter is 2√5 units, so the radius is √5 units and the area is 5π square units.

🪂 Understanding Parabolas

🪂 Definition of a parabola

A parabola is the set of points in a plane equidistant from a given line (the directrix) and a point not on the line (the focus).

Every point on a parabola has the same distance to the focus as to the directrix.
The vertex is the point on the parabola where the distance to the directrix is minimum.
Parabolas are also formed by intersecting a cone with a plane parallel to the side of the cone.

🔄 Four orientations of parabolas

Opens	Quadratic in	Function?	Leading coefficient
Upward	x (y = ax² + ...)	Yes	a > 0
Downward	x (y = ax² + ...)	Yes	a < 0
Right	y (x = ay² + ...)	No	a > 0
Left	y (x = ay² + ...)	No	a < 0

Parabolas opening up or down are functions (pass the vertical line test).
Parabolas opening left or right are not functions because they fail the vertical line test.
Don't confuse: the variable that is squared determines the orientation—if y² appears, the parabola opens horizontally.

📐 General Form vs Standard Form

📐 Two ways to write parabola equations

General form:

y = ax² + bx + c (opens up/down)
x = ay² + by + c (opens left/right)

Standard form (vertex form):

y = a(x − h)² + k (opens up/down)
x = a(y − k)² + h (opens left/right)

Where a, b, and c are real numbers and a ≠ 0.

🎯 Why standard form matters

Standard form immediately reveals the vertex: (h, k).
The vertex is the turning point of the parabola.
Example: In y = (x + 3)² + 2, rewrite as y = [x − (−3)]² + 2, so the vertex is (−3, 2).
This represents a horizontal shift left 3 units and vertical shift up 2 units from the basic parabola y = x².

🔨 Completing the Square

🔨 The technique

Completing the square transforms general form into standard form by creating a perfect square trinomial.

Steps:

If the leading coefficient is not 1, factor it out from the first two terms.
Inside the parentheses, add and subtract (b/2)² where b is the coefficient of the linear term.
Factor the perfect square trinomial.
Distribute any factored-out coefficient and simplify.

🔨 Example with leading coefficient 1

Starting with y = x² − 8x + 15:

Identify b = −8, so (b/2)² = (−8/2)² = 16.
Rewrite: y = (x² − 8x + 16) + 15 − 16.
Factor: y = (x − 4)² − 1.
Vertex: (4, −1).

🔨 Example with leading coefficient ≠ 1

Starting with y = −2x² + 12x − 16:

Factor out −2 from first two terms: y = −2(x² − 6x) − 16.
Inside parentheses, b = −6, so (b/2)² = 9.
Add and subtract 9: y = −2(x² − 6x + 9 − 9) − 16.
Factor: y = −2[(x − 3)² − 9] − 16.
Distribute: y = −2(x − 3)² + 18 − 16 = −2(x − 3)² + 2.
Vertex: (3, 2).

Important: Adding and subtracting the same value is equivalent to adding 0, so it doesn't change the expression.

📊 Graphing Parabolas

📊 Using both forms together

General form helps find intercepts easily (set x = 0 for y-intercept, y = 0 for x-intercept).
Standard form reveals the vertex (h, k) and opening direction.
Combine information from both forms for a complete sketch.

📊 Graphing parabolas that open up or down

Example: y = −2x² + 12x − 16 (from earlier, standard form: y = −2(x − 3)² + 2)

Key features:

Leading coefficient a = −2 < 0, so opens downward.
Vertex: (3, 2) from standard form.
y-intercept: set x = 0 in general form → y = −16, so (0, −16).
x-intercepts: set y = 0 in standard form → 0 = −2(x − 3)² + 2 → (x − 3)² = 1 → x = 2 or x = 4, giving (2, 0) and (4, 0).

📊 Graphing parabolas that open left or right

Example: x = y² + 10y + 13

Key features:

Quadratic in y with a = 1 > 0, so opens to the right.
x-intercept: set y = 0 → x = 13, so (13, 0).
Complete the square: x = (y² + 10y + 25) − 25 + 13 = (y + 5)² − 12.
Vertex: (−12, −5) (remember h and k placement for horizontal parabolas).
y-intercepts: set x = 0 → (y + 5)² = 12 → y = −5 ± 2√3.

Don't confuse: For x = a(y − k)² + h, the vertex is (h, k), not (k, h)—the h value is still the x-coordinate.

📊 When there are no intercepts

A parabola opening left with vertex to the left of the y-axis has no y-intercepts.
A parabola opening down with vertex below the x-axis has no x-intercepts.
In such cases, choose additional points by selecting values for the non-squared variable and solving for the other.

Circles

🧭 Overview

🧠 One-sentence thesis

A circle is completely determined by its center and radius, and its equation can be written in standard form to reveal these properties directly or converted from general form by completing the square.

📌 Key points (3–5)

Definition: A circle is the set of all points in a plane at a fixed distance (radius) from a center point.
Standard form reveals structure: The equation (x - h)² + (y - k)² = r² immediately shows center (h, k) and radius r.
General form requires conversion: When given x² + y² + cx + dy + e = 0, complete the square for both x and y to find the center and radius.
Common confusion: Don't mix up the signs—(x - 2)² means the center x-coordinate is +2, not -2; watch for subtraction inside the parentheses.
The unit circle: The special case x² + y² = 1 has center (0, 0) and radius 1, and can be split into two functions for the top and bottom halves.

📐 Circle definition and standard form

📐 What a circle is

A circle is the set of points in a plane that lie a fixed distance from any point, called the center.

The radius is the fixed distance from the center to any point on the circle.
The diameter is the length of a line segment passing through the center with endpoints on the circle.
A circle can be formed by intersecting a cone with a plane perpendicular to the cone's axis.

📝 Standard form equation

The equation of a circle in standard form is (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius.

This form comes from applying the distance formula between the center (h, k) and any point (x, y) on the circle, then squaring both sides.
How to read it:
- The center coordinates are (h, k)
- The radius is r (take the square root of the right side if needed)
Example: (x - 2)² + (y + 5)² = 16 has center (2, -5) and radius r = 4

🔍 Reading the signs correctly

Don't confuse: The signs inside the parentheses are opposite to the center coordinates.

Equation term	Center coordinate
(x - 3)²	h = 3
(x + 4)²	h = -4
(y - 4)²	k = 4
(y + 2)²	k = -2

When you see (x - h)², the center x-value is +h
When you see (y + 5)², rewrite as [y - (-5)]² to see that k = -5

🎨 Graphing circles

🎨 How to graph from standard form

Step-by-step process:

Identify the center (h, k) from the equation
Identify the radius r (square root of the right side if necessary)
Plot the center point
From the center, mark points r units up, down, left, and right
Sketch the circle through these four points

Example: For (x - 2)² + (y + 5)² = 16, center is (2, -5) and r = 4, so mark points at (2, -1), (2, -9), (-2, -5), and (6, -5), then draw the circle.

🔢 Finding intercepts

For y-intercepts: Set x = 0 and solve for y For x-intercepts: Set y = 0 and solve for x

You may get real solutions (intercepts exist), complex solutions (no intercepts), or no solution.
Example: (x - 2)² + (y + 5)² = 16 has y-intercepts but no x-intercepts because solving for x yields complex numbers.

⭕ The unit circle

The unit circle is the circle centered at the origin with radius 1; its equation is x² + y² = 1.

Center: (0, 0)
Radius: r = 1
Can be solved for y to get two functions:
- y = √(1 - x²) represents the top half
- y = -√(1 - x²) represents the bottom half
This is important for trigonometry and other applications.

🔄 Converting between forms

🔄 General form of a circle

The equation of a circle in general form is x² + y² + cx + dy + e = 0, where c, d, and e are real numbers.

Both x² and y² have the same coefficient (usually 1 after dividing through).
This form does not immediately reveal the center or radius.
How to recognize it: Look for both x² and y² with equal coefficients.

🔧 Completing the square

To convert from general to standard form:

Group variables: Move the constant to the right side; group x terms together and y terms together
Complete the square for x: Add (coefficient of x / 2)² to both sides
Complete the square for y: Add (coefficient of y / 2)² to both sides
Factor: Write each grouping as a perfect square
Read off: Identify center and radius from the standard form

Example: x² + y² + 6x - 8y + 13 = 0

Group: (x² + 6x + ___) + (y² - 8y + ___) = -13
Complete: (x² + 6x + 9) + (y² - 8y + 16) = -13 + 9 + 16
Factor: (x + 3)² + (y - 4)² = 12
Result: Center (-3, 4), radius r = √12 = 2√3

⚠️ When coefficients aren't 1

If the equation has coefficients other than 1 on x² and y² (e.g., 4x² + 4y²), first divide the entire equation by that coefficient to get standard form with coefficient 1.

Example: 4x² + 4y² - 8x + 12y - 3 = 0 becomes x² + y² - 2x + 3y - 3/4 = 0 after dividing by 4.

🎯 Key properties and applications

🎯 What determines a circle

A circle is completely determined by:

Its center point (h, k)
Its radius r

Given any two of these pieces of information (or equivalent information like "passing through a point" or "diameter endpoints"), you can write the equation.

📏 Finding equations from descriptions

Given center and radius: Substitute directly into (x - h)² + (y - k)² = r²

Given center and a point on the circle:

Use the distance formula to find r
Then write the equation

Given diameter endpoints:

Find the center using the midpoint formula
Find the radius as half the distance between endpoints

🔍 Recognizing circle equations

How to tell a circle from other conic sections:

Both x² and y² appear
Both have the same coefficient
No xy term (that would be a rotated conic)
Don't confuse with a parabola, which has only x² or only y², not both

Example: x² + y² + 6x - 8y + 13 = 0 is a circle (both squared terms, same coefficient)

Example: y = x² + 5x - 1 is a parabola (only x is squared)

Ellipses

🧭 Overview

🧠 One-sentence thesis

An ellipse is determined by its center, major radius, minor radius, and orientation, all of which can be read from its standard-form equation to enable graphing and analysis.

📌 Key points (3–5)

What an ellipse is: the set of points whose distances from two fixed points (foci) sum to a constant; graphically, an oval with a longest diameter (major axis) and shortest diameter (minor axis).
Standard form reveals structure: the equation (x − h)²/a² + (y − k)²/b² = 1 shows center (h, k), and the larger of a or b determines the major radius and orientation.
Orientation depends on a vs b: if a > b the ellipse is horizontal (wider); if a < b it is vertical (taller).
Common confusion: distinguishing major vs minor radius—always compare a and b; the larger value corresponds to the major radius, regardless of which variable it's attached to.
Converting from general form: group like terms, complete the square for both x and y, then divide to get 1 on the right side.

📐 Defining the ellipse

📐 Geometric definition

An ellipse is the set of points in a plane whose distances from two fixed points (foci) have a sum equal to a positive constant.

If F₁ and F₂ are the foci and d is the constant, then a point (x, y) is on the ellipse when d = d₁ + d₂ (the sum of distances to each focus).
An ellipse can also be formed by slicing a cone with an oblique plane that does not intersect the base.

📏 Key parts of an ellipse

Term	Definition
Vertices	Points on the ellipse at the endpoints of the major axis (maximum distance apart)
Center	Midpoint between the vertices
Major axis	Line segment through the center where distance between endpoints is maximum
Minor axis	Line segment through the center where distance between endpoints is minimum
Co-vertices	Endpoints of the minor axis
Major radius (a)	One-half the length of the major axis
Minor radius (b)	One-half the length of the minor axis

📝 Standard form equation

📝 The standard form

(x − h)²/a² + (y − k)²/b² = 1

Center: (h, k)
The larger of a and b is the major radius; the smaller is the minor radius.
Critical requirement: the equation must equal 1 (not any other number).

🧭 Determining orientation

Horizontal ellipse: a > b; the major axis is parallel to the x-axis; vertices are at (h ± a, k).
Vertical ellipse: a < b; the major axis is parallel to the y-axis; vertices are at (h, k ± b).
Special case: if a = b, the ellipse becomes a circle.

Example: For (x − 1)²/4 + (y − 8)²/9 = 1, center is (1, 8), a = 2, b = 3. Since b > a, the ellipse is vertical.

🖊️ Graphing from standard form

Identify the center (h, k).
Mark points a units left and right from the center.
Mark points b units up and down from the center.
Sketch an ellipse through these four points.

Don't confuse: The values a and b come from taking square roots of the denominators (a² and b²), not directly from the equation.

🔄 Converting from general form

🔄 General form structure

px² + qy² + cx + dy + e = 0, where p, q > 0

Both x² and y² appear with positive coefficients.
If the coefficients are equal (p = q), the equation represents a circle, not an ellipse.

🔧 Conversion steps

Group and factor: Collect x terms together, y terms together; move the constant to the right; factor so each squared term has coefficient 1.
Complete the square: For each variable grouping, add the appropriate constant inside parentheses; balance the equation by accounting for the factor outside.
Divide to get 1: Divide both sides so the right side equals 1.
Read the parameters: Identify center (h, k), a, and b from the standard form.

Example: Starting with 2x² + 9y² + 16x − 90y + 239 = 0:

Group: 2(x² + 8x) + 9(y² − 10y) = −239
Complete the square: 2(x² + 8x + 16) + 9(y² − 10y + 25) = −239 + 32 + 225
Simplify: 2(x + 4)² + 9(y − 5)² = 18
Divide by 18: (x + 4)²/9 + (y − 5)²/2 = 1
Result: center (−4, 5), a = 3, b = √2, horizontal orientation.

⚖️ Balancing when completing the square

When you add a constant inside a factored grouping, you must multiply that constant by the factor to find what you're adding to the right side.
Example: Adding 16 inside 2(x² + 8x + ___) means you're really adding 2 × 16 = 32 to the equation.

🔍 Finding intercepts

🔍 x-intercepts

Set y = 0 in the equation and solve for x.
May result in real solutions, complex solutions (meaning no x-intercepts), or no solutions.

🔍 y-intercepts

Set x = 0 in the equation and solve for y.
Similarly, may or may not yield real intercepts.

Example: For (x + 3)²/4 + (y − 2)²/25 = 1:

x-intercepts: Set y = 0, solve to get x = (−15 ± 2√21)/5
y-intercepts: Set x = 0, leads to complex solutions, so none exist.

Don't confuse: Not all ellipses have both x- and y-intercepts; it depends on the position of the center relative to the axes.

📊 Additional properties

📊 Area of an ellipse

Formula: A = πab, where a and b are the major and minor radii.
Example: If a = 5 and b = √5, then area = 5π√5 square units.

🔗 Relationship to functions

An ellipse centered at the origin can be split into two functions by solving for y.
Example: x² + y²/4 = 1 gives y = ±2√(1 − x²); the positive root is the top half, the negative root is the bottom half.

Hyperbolas

🧭 Overview

🧠 One-sentence thesis

A hyperbola is defined by two fixed points (foci) such that the absolute difference of distances from any point on the curve to the two foci is constant, and its graph is completely determined by its center, vertices, and asymptotes.

📌 Key points (3–5)

What a hyperbola is: the set of points where the absolute difference of distances to two foci equals a positive constant; consists of two separate curves called branches.
Two standard forms: hyperbolas open left-right when the x-term is positive, or up-down when the y-term is positive.
How to graph: use the center (h, k), mark points a units and b units away, draw the fundamental rectangle, sketch asymptotes through its corners, then draw branches through the vertices.
Common confusion: in the upward-downward form, the y-term comes first but the y-value of the center still comes from the quantity involving y (not automatically the first coordinate).
Identifying hyperbolas: quadratic in both x and y with coefficients of opposite signs.

📐 Definition and structure

📐 What defines a hyperbola

A hyperbola is the set of points in a plane whose distances from two fixed points, called foci, has an absolute difference that is equal to a positive constant.

If F₁ and F₂ are the foci and d is a positive constant, then (x, y) is on the hyperbola if d = |d₁ − d₂|.
A hyperbola is also formed by the intersection of a cone with an oblique plane that intersects the base.
It consists of two separate curves called branches.

🎯 Key components

Component	Definition
Vertices	Points on the separate branches where the distance is at a minimum
Center	The midpoint between the vertices
Transverse axis	The line segment formed by the vertices
Conjugate axis	A line segment through the center perpendicular to the transverse axis
Fundamental rectangle	The rectangle formed using the endpoints of the transverse and conjugate axes
Asymptotes	Lines through the corners of the fundamental rectangle with slopes ±b/a; they guide the shape but are not part of the graph

📝 Standard forms

📝 Left-right opening hyperbola

Standard form: (x − h)² / a² − (y − k)² / b² = 1

The center is (h, k).
The vertices are (h ± a, k).
The x-term has the positive coefficient.
Example: If the expression involving x is positive, the hyperbola opens left and right.

📝 Up-down opening hyperbola

Standard form: (y − k)² / b² − (x − h)² / a² = 1

The center is (h, k).
The vertices are (h, k ± b).
The y-term has the positive coefficient.
Don't confuse: Even though the y-term appears first, the y-value of the center still comes from the quantity involving y, and the x-value from the quantity involving x.

🔍 How to tell which way it opens

If the coefficient of the x² term is positive (and y² is negative), the hyperbola opens left and right.
If the coefficient of the y² term is positive (and x² is negative), the hyperbola opens upward and downward.

🎨 Graphing procedure

🎨 Step-by-step graphing from standard form

Identify the center (h, k) from the equation.
Find a and b: a is the square root of the denominator under the positive term; b is the square root of the other denominator.
Mark the fundamental rectangle: from the center, mark points a units left and right, and b units up and down; connect these to form a rectangle.
Draw asymptotes: sketch dashed lines through opposite corners of the rectangle (slopes are ±b/a).
Draw the branches:
- If x² is positive, draw curves opening left and right through the vertices at (h ± a, k).
- If y² is positive, draw curves opening up and down through the vertices at (h, k ± b).

Example: For (x − 5)² / 9 − (y − 4)² / 4 = 1, the center is (5, 4), a = 3, b = 2, and it opens left-right.

📍 Finding intercepts

x-intercepts: set y = 0 and solve for x.
y-intercepts: set x = 0 and solve for y.
Not all hyperbolas have both types of intercepts; some may have none of one type.

🔄 Converting from general form

🔄 General form

General form: px² − qy² + cx + dy + e = 0 (opens left-right) or qy² − px² + cx + dy + e = 0 (opens up-down), where p, q > 0.

🔄 Conversion steps

Group terms with the same variables and move the constant to the right side.
Factor so the leading coefficient of each grouping is 1.
Complete the square for both x and y groupings; remember that the factor in front affects the balancing value on the right side.
Divide to obtain 1 on the right side.
Read off the center, a, and b from the resulting standard form.

Example: Starting with 4x² − 9y² + 32x − 54y − 53 = 0, group and factor to get 4(x² + 8x) − 9(y² + 6y) = 53, then complete the square for each grouping.

🔍 Identifying conic sections

🔍 Recognition rules

Conic	Characteristic
Parabola	Quadratic in only one variable, linear in the other
Circle	Quadratic in both variables; coefficients of squared terms are the same
Ellipse	Quadratic in both variables; coefficients of squared terms are different but have the same sign
Hyperbola	Quadratic in both variables; coefficients of squared terms have different (opposite) signs

🔍 Hyperbola identification

Look for x² and y² terms with opposite signs (one positive, one negative).
General forms: px² − qy² + cx + dy + e = 0 or qy² − px² + cx + dy + e = 0, where p, q > 0.
The term with the positive coefficient determines the opening direction.

Solving Nonlinear Systems

🧭 Overview

🧠 One-sentence thesis

Nonlinear systems—where at least one equation is not linear—can be solved using the substitution method to find ordered pairs that satisfy all equations simultaneously.

📌 Key points (3–5)

What a nonlinear system is: a system of equations where at least one equation is not linear.
Solution format: solutions are ordered pairs (x, y) that satisfy both equations, just as in linear systems.
Method used: the substitution method—solve for one variable in one equation, substitute into the other, then solve.
Common confusion: nonlinear systems often yield multiple solutions (e.g., two y-values), unlike typical linear systems that have one solution.
Key difference from linear systems: the presence of squared terms or other nonlinear expressions changes the solution process and the number of solutions.

🔍 What makes a system nonlinear

🔍 Definition and identification

Nonlinear system: a system of equations where at least one equation is not linear.

"Not linear" means the equation contains terms like x squared, y squared, xy, or other nonlinear expressions.
In the excerpt's example, the system has two equations: x + 2y = 0 (linear) and x squared + y squared = 5 (nonlinear).
Because at least one equation is nonlinear, the entire system is classified as nonlinear.

🎯 What solutions look like

Solutions are still ordered pairs (x, y).
These pairs must satisfy both equations in the system.
Don't confuse: the format of solutions is the same as for linear systems, but the process and number of solutions differ.

🔄 Solving with substitution

🔄 The substitution method

The excerpt states: "we will use the substitution method to solve nonlinear systems."
Steps illustrated in Example 1:
1. Solve for one variable in one equation (here, solve for x in the first equation).
2. Substitute that expression into the other equation.
3. Solve for the remaining variable.

🧮 Example walkthrough

Example: The system is x + 2y = 0 and x squared + y squared = 5.

Step 1: Solve the first equation for x → x = −2y.
Step 2: Substitute x = −2y into the second equation (x squared + y squared = 5).
Step 3: Solve for y. The excerpt notes "there are two answers for y," indicating multiple solutions.

⚠️ Multiple solutions

The excerpt emphasizes that "there are two answers for y."
This is typical in nonlinear systems because squared terms can produce two values.
Don't confuse: linear systems usually have one solution (or none, or infinitely many if dependent); nonlinear systems often have two or more distinct solutions.

📊 Key characteristics summary

Aspect	What the excerpt says
Definition	At least one equation is not linear
Solution format	Ordered pairs (x, y) that satisfy both equations
Method	Substitution method
Typical outcome	Multiple solutions (e.g., two y-values in the example)

Introduction to Sequences and Series

🧭 Overview

🧠 One-sentence thesis

Sequences are ordered lists generated by formulas, and series are the sums of sequence terms, both of which can be expressed compactly using sigma notation.

📌 Key points (3–5)

What a sequence is: a function whose domain is consecutive natural numbers starting with 1, producing an ordered list of terms.
General term notation: the nth term is written as a_n (read "a sub n"), found by substituting values of n into the formula.
Sequence vs series: a sequence is an ordered list; a series is the sum of the terms in that list.
Common confusion: infinite sequence vs finite sequence—infinite sequences continue forever (domain {1, 2, 3, …}), while finite sequences end at some natural number k.
Sigma notation: a compact way to write series using the symbol Σ, specifying the index range and the general term.

📐 Understanding Sequences

📐 What defines a sequence

Sequence: a function whose domain is a set of consecutive natural numbers beginning with 1.

The domain determines whether the sequence is infinite or finite.
Infinite sequence: domain is {1, 2, 3, …} and continues forever.
Finite sequence: domain is {1, 2, 3, …, k} where k is a natural number, so it stops.
The output values are called terms of the sequence.
Order matters (unlike sets).

🔢 General term and subscript notation

General term of a sequence: an equation that defines the nth term, commonly denoted using subscripts a_n.

To find specific terms, substitute n = 1, 2, 3, etc. into the formula.
Example: if a_n = 5n − 3, then:
- First term (n=1): a_1 = 5(1) − 3 = 2
- Second term (n=2): a_2 = 5(2) − 3 = 7
- Third term (n=3): a_3 = 5(3) − 3 = 12
This produces the ordered list: 2, 7, 12, 17, 22, …
The ellipsis (…) indicates the sequence continues.

🔄 Alternating signs and other variables

Some sequences alternate in sign using powers of (−1).
Example: a_n = (−1)^n x^(n+1) produces −x², x³, −x⁴, x⁵, −x⁶, …
The variable n is the index; other variables (like x) remain as symbols.
Don't confuse: replace only the index variable (n), not other variables in the formula.

🔁 Recurrence relations

Recurrence relation: a formula that describes a sequence in terms of its previous terms.

Instead of a direct formula for a_n, you define a_n using earlier terms like a_(n−1) or a_(n−2).
Fibonacci sequence is a classic example:
- F_1 = 1, F_2 = 1
- F_n = F_(n−2) + F_(n−1) for n > 2
- Each term is the sum of the two preceding terms: 1, 1, 2, 3, 5, 8, 13, …
The Fibonacci sequence appears in art, computer science, biology, and the Fibonacci spiral found in nature.

➕ Understanding Series

➕ What a series is

Series: the sum of the terms of a sequence.

A sequence is a list; a series is the sum of that list.
Infinite series (denoted S_∞): the sum of all terms in an infinite sequence.
Example: S_∞ = 1 + 3 + 5 + 7 + 9 + ⋯ (sum of positive odd integers).

🧮 Partial sums

Partial sum (denoted S_n): the sum of the first n terms in a sequence.

S_n adds only the first n terms, not the entire infinite sequence.
Example: for the sequence 1, 3, 5, 7, 9, …
- S_5 = 1 + 3 + 5 + 7 + 9 = 25 (5th partial sum)
Example: for the sequence 3, −6, 12, −24, 48, …
- S_3 = 3 + (−6) + 12 = 9
- S_5 = 3 + (−6) + 12 + (−24) + 48 = 33

🔣 Sigma Notation

🔣 What sigma notation means

Sigma notation (or summation notation): a compact way to denote a series using the symbol Σ (upper case Greek letter sigma).

General form: Σ (from k=1 to n) a_k = a_1 + a_2 + ⋯ + a_n
The index of summation (often k, n, or i) is the variable that changes.
The number below Σ is the starting integer; the number above is the ending integer.
Replace n with ∞ to indicate an infinite series.

📝 How to expand sigma notation

Substitute each integer value of the index into the general term and add the results.
Example: Σ (from k=1 to 5) (−3)^(k−1)
- k=1: (−3)^0 = 1
- k=2: (−3)^1 = −3
- k=3: (−3)^2 = 9
- k=4: (−3)^3 = −27
- k=5: (−3)^4 = 81
- Sum: 1 − 3 + 9 − 27 + 81 = 61
The index does not always start at 1; it can start at 0, 2, or any integer.

♾️ Infinite series in sigma notation

Use ∞ as the upper bound to indicate the series continues forever.
Example: Σ (from n=0 to ∞) n/(n+1) expands to 0 + 1/2 + 2/3 + 3/4 + ⋯
The ellipsis shows the pattern continues indefinitely.

⚠️ Index replacement caution

When expanding, replace only the index variable, not other variables.
Example: Σ (from i=1 to ∞) (−1)^(i−1) x^(2i) expands to x² − x⁴ + x⁶ − ⋯
The variable x remains; only i is substituted with 1, 2, 3, …
Don't confuse: the index (i, k, n) is the counter; other letters are parameters or variables in the expression.

🔑 Key Takeaways Summary

Concept	Definition	Notation
Sequence	Ordered list from a function on natural numbers	a_n (nth term)
Infinite sequence	Domain is {1, 2, 3, …}	Continues forever
Finite sequence	Domain is {1, 2, 3, …, k}	Stops at k
Series	Sum of terms in a sequence	S_∞ or S_n
Partial sum	Sum of first n terms	S_n
Sigma notation	Compact summation symbol	Σ (from k=start to end) a_k
Recurrence relation	Formula using previous terms	Example: F_n = F_(n−2) + F_(n−1)

Arithmetic Sequences and Series

🧭 Overview

🧠 One-sentence thesis

Arithmetic sequences grow by adding a constant difference to each term, and their sums can be calculated efficiently using formulas based on the first term, last term, and number of terms.

📌 Key points (3–5)

What defines an arithmetic sequence: each term equals the previous term plus a constant difference d.
General term formula: any term can be found using a_n = a_1 + (n - 1)d, where a_1 is the first term and d is the common difference.
Arithmetic series formula: the sum of the first n terms is S_n = n(a_1 + a_n) / 2, using only the first term, last term, and count.
Common confusion: the common difference can be negative (sequence decreases) or positive (sequence increases); always check by subtracting successive terms.
Why it matters: these formulas let you find distant terms or large sums without tedious addition (e.g., sum of first 100 integers).

🔢 What is an arithmetic sequence

🔢 Definition and structure

Arithmetic sequence (or arithmetic progression): a sequence of numbers where each successive number is the sum of the previous number and some constant d.

The constant d is called the common difference.
You find d by subtracting any term from the next: d = a_n - a_(n-1).
Example: 1, 3, 5, 7, 9, … has d = 2 (each term is 2 more than the previous).

🧮 Building the sequence step-by-step

The excerpt shows how each term is constructed:

Start with a_1.
a_2 = a_1 + d
a_3 = a_2 + d = a_1 + 2d
a_4 = a_3 + d = a_1 + 3d
In general, a_n = a_1 + (n - 1)d.

This pattern holds because you add d exactly (n - 1) times to reach the nth term.

➕ Positive vs ➖ negative common difference

Positive d: sequence increases (e.g., 7, 10, 13, 16, …).
Negative d: sequence decreases (e.g., 6, 4, 2, 0, -2, …).
Don't confuse: a negative d still follows the same formula; just substitute the negative value.

Example from the excerpt: for 6, 4, 2, 0, -2, … the common difference is d = 4 - 6 = -2, so a_n = 6 + (n - 1)(-2) = 8 - 2n.

🧩 Finding the general term

🧩 Using the formula a_n = a_1 + (n - 1)d

To write an equation for the nth term:

Identify the first term a_1.
Calculate the common difference d.
Substitute into a_n = a_1 + (n - 1)d and simplify.

Example: for 7, 10, 13, 16, 19, …

a_1 = 7, d = 3.
a_n = 7 + (n - 1) · 3 = 7 + 3n - 3 = 3n + 4.
The 100th term is a_100 = 3(100) + 4 = 304.

🔍 When the first term is not given

If you know two non-consecutive terms (e.g., a_3 = -1 and a_10 = 48), set up a system:

a_3 = a_1 + 2d = -1
a_10 = a_1 + 9d = 48

Subtract the first equation from the second to eliminate a_1:

7d = 49, so d = 7.
Substitute back: a_1 + 14 = -1, so a_1 = -15.
General term: a_n = -15 + (n - 1) · 7 = 7n - 22.

📐 Arithmetic means

Arithmetic means: the terms between given terms of an arithmetic sequence.

Example: to find all terms between a_1 = -8 and a_7 = 10:

Use a_7 = a_1 + 6d → 10 = -8 + 6d → d = 3.
Generate a_2, a_3, …, a_6 using a_n = 3n - 11: the means are -5, -2, 1, 4, 7.

➕ Arithmetic series and sums

➕ What is an arithmetic series

Arithmetic series: the sum of the terms of an arithmetic sequence.

Example: the sum of the first 5 terms of a_n = 2n - 1 is 1 + 3 + 5 + 7 + 9 = 25.
For large n, adding term-by-term is impractical.

🧮 Deriving the sum formula

The excerpt derives the nth partial sum formula by a clever trick:

Write the sum forward: S_n = a_1 + (a_1 + d) + (a_1 + 2d) + … + a_n.
Write the sum backward: S_n = a_n + (a_n - d) + (a_n - 2d) + … + a_1.
Add the two equations: each pair sums to (a_1 + a_n), and there are n pairs.
Result: 2S_n = n(a_1 + a_n).
Divide by 2: S_n = n(a_1 + a_n) / 2.

This formula requires only the first term, last term, and number of terms.

📊 Using the sum formula

Example: sum of the first 100 positive odd integers.

The sequence is a_n = 2n - 1, so a_1 = 1 and a_100 = 199.
S_100 = 100(1 + 199) / 2 = 100 · 200 / 2 = 10,000.

Example: find the sum of the first 50 terms of 4, 9, 14, 19, 24, …

d = 5, a_1 = 4, a_n = 5n - 1.
a_50 = 5(50) - 1 = 249.
S_50 = 50(4 + 249) / 2 = 50 · 253 / 2 = 6,325.

🔢 Summation notation

When given a sum like Σ(n=1 to 35) (10 - 4n):

Recognize a_n = 10 - 4n as the general term.
Find a_1 = 10 - 4(1) = 6 and a_35 = 10 - 4(35) = -130.
S_35 = 35(6 + (-130)) / 2 = 35(-124) / 2 = -2,170.

🎭 Real-world application

🎭 Theater seating problem

The excerpt gives an example: an amphitheater has 26 seats in the first row, 28 in the second, 30 in the third, and so on for 18 rows. Find total capacity.

The sequence is 26, 28, 30, … with d = 2.
General term: a_n = 26 + (n - 1) · 2 = 2n + 24.
a_18 = 2(18) + 24 = 60.
Total seats: S_18 = 18(26 + 60) / 2 = 18 · 86 / 2 = 774.

Don't confuse: you need the sum (total seats), not just the last term (seats in row 18).

🔑 Key takeaways from the excerpt

The excerpt emphasizes:

Identifying the common difference is the first step; check by subtracting successive terms.
The general term formula a_n = a_1 + (n - 1)d works for any arithmetic sequence, even when d is negative or fractional.
The sum formula S_n = n(a_1 + a_n) / 2 avoids tedious addition and scales to large n.
Linear in n: any general term that is linear (e.g., a_n = 3n + 4) defines an arithmetic sequence.
Applications: real-world problems (seating, salary contracts, stacking) often involve arithmetic sequences and series.

Geometric Sequences and Series

🧭 Overview

🧠 One-sentence thesis

A geometric sequence is built by multiplying each term by a constant ratio, and any such sequence can be expressed using its first term, common ratio, and position index.

📌 Key points (3–5)

What defines a geometric sequence: each successive number is the product of the previous number and a constant r (the common ratio).
How to find the common ratio: divide any term by the previous term; the result r is constant throughout the sequence.
General formula: any geometric sequence can be written as a_n = a_1 × r^(n−1), where a_1 is the first term, r is the common ratio, and n is the position.
Common confusion: the exponent is (n−1), not n, because the first term a_1 already has no multiplication by r.
Geometric means: the terms that fall between two given terms of a geometric sequence.

🔢 Core definition and structure

🔢 What is a geometric sequence

Geometric sequence (or geometric progression): a sequence of numbers where each successive number is the product of the previous number and some constant r.

The relationship is: a_n = r × a_(n−1).
The constant r is obtained by dividing any term by the term before it: a_n / a_(n−1) = r.
This constant is called the common ratio.

🧮 The common ratio

Common ratio: the constant r obtained from dividing any two successive terms of a geometric sequence.

It is the same for every pair of consecutive terms.
Example from the excerpt: the sequence 9, 27, 81, 243, 729… has a common ratio of 3 because 27/9 = 3, 81/27 = 3, and so on.
The ratio tells you how much to multiply each term to get the next one.

🧩 Building the general term

🧩 Step-by-step construction

The excerpt shows how to express any term in terms of the first term and the common ratio:

a_1 = a_1 (the first term, no multiplication yet)
a_2 = r × a_1
a_3 = r × a_2 = r × (a_1 × r) = a_1 × r²
a_4 = r × a_3 = r × (a_1 × r²) = a_1 × r³
a_5 = r × a_4 = r × (a_1 × r³) = a_1 × r⁴

Notice the pattern: the exponent on r is always one less than the term's position.

📐 The general formula

General term of a geometric sequence: a_n = a_1 × r^(n−1)

a_1 is the first term.
r is the common ratio.
n is the position (index) of the term.
The exponent is (n−1) because the first term has not been multiplied by r yet.
Any general term that is exponential in n is a geometric sequence.

🔍 Don't confuse the exponent

The formula is r^(n−1), not r^n.
Why: when n = 1, you want a_1 × r^0 = a_1 (the first term unchanged).
Example: for the 10th term, use r^(10−1) = r^9, not r^10.

🧪 Working with examples

🧪 Finding the general term and a specific term

The excerpt provides Example 1: the sequence 3, 6, 12, 24, 48…

Step 1: Find the common ratio

r = 6/3 = 2 (check: 12/6 = 2, 24/12 = 2, etc.)

Step 2: Write the general term

a_1 = 3, r = 2
a_n = 3 × (2)^(n−1)

Step 3: Calculate a specific term

For the 10th term: a_10 = 3 × (2)^(10−1) = 3 × 2^9 = 3 × 512 = 1,536

🔗 Geometric means

Geometric means: the terms between given terms of a geometric sequence.

The excerpt provides Example 2: find all terms between a_1 = −5 and a_4 = −135.

Step 1: Use the general formula to find r

a_n = a_1 × r^(n−1)
For n = 4: a_4 = a_1 × r^(4−1) = a_1 × r³
Substitute: −135 = −5 × r³
Solve: r³ = 27, so r = 3

Step 2: Write the general term

a_n = −5 × (3)^(n−1)

Step 3: Find the missing terms

The excerpt shows the setup; the geometric means are a_2 and a_3, which can be calculated using the formula with n = 2 and n = 3.

📊 Summary table

Concept	Definition / Formula	Key point
Geometric sequence	Each term = previous term × r	Constant multiplication
Common ratio r	a_n / a_(n−1)	Same for all consecutive pairs
General term	a_n = a_1 × r^(n−1)	Exponent is (n−1), not n
Geometric means	Terms between two given terms	Found by solving for r first

Geometric Sequences and Series

Binomial Theorem

🧭 Overview

🧠 One-sentence thesis

A geometric sequence is built by multiplying each term by a constant ratio, and its sum (when the ratio is between -1 and 1) can be calculated using formulas that depend on the first term and that common ratio.

📌 Key points (3–5)

What defines a geometric sequence: each term equals the previous term multiplied by a constant ratio r.
General term formula: any term can be written as a_n = a_1 × r^(n-1), using only the first term, the ratio, and the position.
Finite sum formula: the sum of the first n terms is S_n = a_1(1 - r^n)/(1 - r) when r ≠ 1.
Infinite sum condition: if the absolute value of r is less than 1, the infinite series converges to S_∞ = a_1/(1 - r); otherwise no sum exists.
Common confusion: convergent vs divergent—only when |r| < 1 does an infinite geometric series have a finite sum; if |r| ≥ 1, the series grows without bound.

🔢 What is a geometric sequence

🔢 Definition and common ratio

Geometric sequence (or geometric progression): a sequence where each successive number is the product of the previous number and some constant r.

The constant r is called the common ratio because dividing any term by the previous term always gives r.
Example from the excerpt: 9, 27, 81, 243, 729… has first term a_1 = 9 and ratio r = 3 (since 27/9 = 3, 81/27 = 3, etc.).
The recurrence relation is a_n = r × a_(n-1).

📐 General term formula

The excerpt shows that any term can be expressed as:

a_2 = r × a_1
a_3 = r × a_2 = a_1 × r²
a_4 = r × a_3 = a_1 × r³
In general: a_n = a_1 × r^(n-1)

This formula lets you find any term directly without computing all previous terms.

Example from the excerpt: For the sequence 3, 6, 12, 24, 48…, the common ratio is r = 6/3 = 2 and a_1 = 3, so a_n = 3(2)^(n-1). The 10th term is a_10 = 3(2)^9 = 3 × 512 = 1,536.

🔍 Geometric means

Geometric means: the terms between given terms of a geometric sequence.

Example from the excerpt: to find all terms between a_1 = -5 and a_4 = -135, first solve for r using a_4 = a_1 × r³, which gives r = 3. Then calculate a_2 = -15 and a_3 = -45.
Don't confuse: you need at least two known terms (and their positions) to determine both a_1 and r when the first term isn't given.

➕ Geometric series and finite sums

➕ What is a geometric series

Geometric series: the sum of the terms of a geometric sequence.

Example: the sum of the first 5 terms of a_n = 3^(n+1) is S_5 = 9 + 27 + 81 + 243 + 729 = 1,089.
Adding many terms by hand is tedious, so a formula is needed.

🧮 Formula for the nth partial sum

The excerpt derives the formula by writing:

S_n = a_1 + a_1 r + a_1 r² + … + a_1 r^(n-1)
Multiply both sides by r: r S_n = a_1 r + a_1 r² + … + a_1 r^n
Subtract: S_n - r S_n = a_1 - a_1 r^n
Factor and divide: S_n = a_1(1 - r^n)/(1 - r) (valid when r ≠ 1).

Example from the excerpt: For the sequence 4, -8, 16, -32, 64…, the ratio is r = -2 and a_1 = 4. The sum of the first 10 terms is S_10 = 4[1 - (-2)^10]/(1 - (-2)) = 4(1 - 1,024)/3 = 4(-1,023)/3 = -1,364.

♾️ Infinite geometric series

♾️ When does an infinite sum exist?

If the common ratio r is a fraction where |r| < 1 (that is, between -1 and 1), then as n increases, the factor (1 - r^n) approaches 1.
The excerpt illustrates: if r = 1/10, then (1 - r²) = 0.99, (1 - r⁴) = 0.9999, (1 - r⁶) = 0.999999—getting closer to 1.
This is expressed as a limit: lim (n→∞) (1 - r^n) = 1 when |r| < 1.

🔁 Convergent series formula

Convergent geometric series: an infinite geometric series where |r| < 1; its sum is given by S_∞ = a_1/(1 - r).

Derivation: start with S_n = a_1(1 - r^n)/(1 - r) = [a_1/(1 - r)] × (1 - r^n). As n → ∞, (1 - r^n) → 1, so S_∞ = a_1/(1 - r).
Example from the excerpt: For the series 3/2 + 1/2 + 1/6 + 1/18 + 1/54 + …, the ratio is r = (1/2)/(3/2) = 1/3. Since |1/3| < 1, the sum is S_∞ = (3/2)/(1 - 1/3) = (3/2)/(2/3) = 9/4.

⚠️ Divergent series

If |r| ≥ 1, the series diverges and has no sum.
Example from the excerpt: if a_n = (5)^(n-1), then r = 5 and S_∞ = 1 + 5 + 25 + … grows without bound.

🔄 Applications

🔄 Repeating decimals

A repeating decimal can be written as an infinite geometric series whose ratio is a power of 1/10.
Example from the excerpt: 0.181818… = 18/100 + 18/10,000 + 18/1,000,000 + … has first term a_1 = 18/100 and ratio r = 1/100. The sum is S_∞ = (18/100)/(1 - 1/100) = (18/100)/(99/100) = 18/99 = 2/11. So 1.181818… = 1 + 2/11 = 1 2/11.

🏀 Bouncing ball problem

A ball bounces back to two-thirds of the height it fell from; if dropped from 27 feet, approximate the total distance traveled.
Falling distances: 27, 18, 12, … form a geometric series with a_1 = 27 and r = 2/3. Sum: S_∞ = 27/(1 - 2/3) = 27/(1/3) = 81 feet.
Rising distances: 18, 12, 8, … form a geometric series with a_1 = 18 and r = 2/3. Sum: S_∞ = 18/(1/3) = 54 feet.
Total distance: 81 + 54 = 135 feet.

📊 Key takeaways from the excerpt

Concept	Formula / Rule	When it applies
General term	a_n = a_1 × r^(n-1)	Any geometric sequence
nth partial sum	S_n = a_1(1 - r^n)/(1 - r)	r ≠ 1
Infinite sum	S_∞ = a_1/(1 - r)	\|r\| < 1 (convergent)
Divergence	No sum exists	\|r\| ≥ 1

Don't confuse:

A geometric sequence (the list of terms) vs a geometric series (the sum of those terms).
Finite sum (always computable if r ≠ 1) vs infinite sum (only exists when |r| < 1).