Algebra | CBC Senior Secondary Mathematics

Foundations

Number Properties & Laws of Indices

Foundation

Before working with unknowns, you must be fluent with the rules that govern all numbers. These properties hold for all real numbers and are the engine behind every algebraic manipulation you will ever perform. In the CBC curriculum, these are introduced at Lower Secondary (Grade 7–9) and reinforced at Senior Secondary as a launch pad for pure mathematics.

The Three Fundamental Laws

Commutative Laws

a + b = b + a a × b = b × a

Associative Laws

(a + b) + c = a + (b + c) (ab)c = a(bc)

Distributive Law

a(b + c) = ab + ac

Why does this matter? These laws justify every step when you expand brackets, collect like terms, or factorise. You use them automatically — but understanding why they work is the difference between a student who memorises and one who can solve any unfamiliar problem.

Laws of Indices (Exponents)

The laws of indices describe how powers behave under multiplication, division, and other operations. Mastery of these is essential for algebra, calculus, and logarithms.

Complete Index Laws — aᵐ, aⁿ ∈ ℝ, a ≠ 0

Product ruleaᵐ × aⁿ = aᵐ⁺ⁿ

Quotient ruleaᵐ ÷ aⁿ = aᵐ⁻ⁿ

Power rule(aᵐ)ⁿ = aᵐⁿ

Product power(ab)ⁿ = aⁿbⁿ

Zero powera⁰ = 1 (a ≠ 0)

Negative powera⁻ⁿ = 1/aⁿ

Fractional power (root)a^(1/n) = ⁿ√a

Fractional power (general)a^(m/n) = (ⁿ√a)ᵐ

Worked Examples — Laws of Indices

Example 1 Simplify: 2³ × 2⁴ ÷ 2²

Step 1Apply the product rule first: 2³ × 2⁴ = 2^(3+4) = 2⁷

Step 2Apply the quotient rule: 2⁷ ÷ 2² = 2^(7−2) = 2⁵

Step 3Evaluate: 2⁵ = 32

Answer: 32

Example 2 Simplify: (3x²y)³

Step 1Apply (ab)ⁿ = aⁿbⁿ to each factor: 3³ × (x²)³ × y³

Step 23³ = 27; apply power rule to x²: (x²)³ = x^(2×3) = x⁶

Answer: 27x⁶y³

Example 3 Evaluate 8^(2/3)

Step 1Use a^(m/n) = (ⁿ√a)ᵐ. Here a = 8, m = 2, n = 3

Step 2Find the cube root: ∛8 = 2

Step 3Raise to the power 2: 2² = 4

Answer: 4

Example 4 Simplify: (16x⁴ / 81y⁸)^(3/4)

Step 1Apply the fractional power to numerator and denominator separately

Step 216^(3/4) = (⁴√16)³ = 2³ = 8; (x⁴)^(3/4) = x³

Step 381^(3/4) = (⁴√81)³ = 3³ = 27; (y⁸)^(3/4) = y⁶

Answer: 8x³ / 27y⁶

Foundations

Algebraic Expressions & Identities

Foundation

An algebraic expression is any combination of numbers, variables, and arithmetic operations. The key skill at this stage is confidently expanding brackets, collecting like terms, and recognising the standard identities that appear repeatedly in CBC examinations. These identities are not arbitrary — each one is a factored or expanded form of the same mathematical object.

Like and Unlike Terms

Terms are like terms if they have exactly the same variable(s) raised to exactly the same power(s). Only like terms can be combined by addition or subtraction.

✓ Like terms (can combine)

3x + 5x = 8x

7x²y − 2x²y = 5x²y

✗ Unlike terms (cannot combine)

3x + 5y → leave as is

2x² + 3x → leave as is

The Standard Algebraic Identities

These identities are used across every topic in Senior Secondary mathematics. Recognising them instantly makes factorisation, expansion, and equation solving significantly faster.

IdentityExpanded Form

(a + b)²a² + 2ab + b²

(a − b)²a² − 2ab + b²

(a + b)(a − b)a² − b² Difference of two squares

(a + b)³a³ + 3a²b + 3ab² + b³

(a − b)³a³ − 3a²b + 3ab² − b³

a³ + b³(a + b)(a² − ab + b²) Sum of cubes

a³ − b³(a − b)(a² + ab + b²) Difference of cubes

Worked Examples — Expressions & Identities

Example 1Expand and simplify: 3(2x + 4) − 2(x − 3)

Step 1Expand the first bracket: 3(2x + 4) = 6x + 12

Step 2Expand the second bracket, taking care with the negative sign: −2(x − 3) = −2x + 6

Step 3Collect like terms: (6x − 2x) + (12 + 6) = 4x + 18

Answer: 4x + 18

Example 2Expand: (x + 3)²

Step 1Use the identity (a + b)² = a² + 2ab + b² with a = x, b = 3

Step 2a² = x², 2ab = 2(x)(3) = 6x, b² = 9

Answer: x² + 6x + 9

Example 3Factorise: 25x² − 16

Step 1Recognise this as a difference of two squares: a² − b²

Step 2Identify: a = 5x (since (5x)² = 25x²) and b = 4 (since 4² = 16)

Step 3Apply: (a + b)(a − b) = (5x + 4)(5x − 4)

Answer: (5x + 4)(5x − 4)

Example 4Factorise: x³ − 8

Step 1Recognise as a difference of cubes: a³ − b³, with a = x, b = 2 (since 2³ = 8)

Step 2Apply: a³ − b³ = (a − b)(a² + ab + b²)

Step 3Substitute: (x − 2)(x² + 2x + 4)

Answer: (x − 2)(x² + 2x + 4)

Core Secondary

Linear & Simultaneous Equations

Secondary

A linear equation in one unknown contains no powers higher than 1. The golden rule of solving any equation is to perform the same operation on both sides. Simultaneous equations extend this to systems of two or more equations with two or more unknowns — a skill that models real-world problems in economics, physics, and engineering.

Standard linear form

ax + b = c ⟹ x = (c − b) / a (a ≠ 0)

Solving Linear Equations — Method

Eliminate fractions by multiplying through by the LCD (if present)
Expand all brackets
Collect all variable terms on one side
Collect all constant terms on the other side
Divide both sides by the coefficient of the variable
Check: substitute your answer back into the original equation

Worked Examples — Linear Equations

Example 1Solve: 3x + 7 = 19

Step 1Subtract 7 from both sides: 3x = 12

Step 2Divide both sides by 3: x = 4

Check3(4) + 7 = 12 + 7 = 19 ✓

x = 4

Example 2Solve: (2x + 1)/3 − (x − 2)/4 = 3

Step 1Multiply every term by the LCD = 12: 4(2x + 1) − 3(x − 2) = 36

Step 2Expand: 8x + 4 − 3x + 6 = 36

Step 3Collect: 5x + 10 = 36 → 5x = 26

Step 4x = 26/5 = 5.2

x = 26/5

Example 3 — Word ProblemA number is doubled and increased by 5, giving 21. Find the number.

Step 1Let the number = x. Write the equation: 2x + 5 = 21

Step 2Subtract 5: 2x = 16 → x = 8

The number is 8

Simultaneous Equations

When two equations have two unknowns, they form a simultaneous system. The three standard methods are elimination, substitution, and the matrix method. All three are CBC-examinable.

Method 1 — Elimination

Make the coefficient of one variable the same in both equations, then add or subtract the equations to eliminate it.

Method 2 — Substitution

Rearrange one equation to express one variable in terms of the other, then substitute into the second equation.

Method 3 — Graphical

Plot both lines on the same axes; the intersection point gives the solution. Useful for verification.

Worked Examples — Simultaneous Equations

Example 4 — EliminationSolve: 2x + 3y = 13 and 4x − y = 5

Step 1Label: equation (1) is 2x + 3y = 13; equation (2) is 4x − y = 5

Step 2Multiply (2) by 3 to match the y-coefficient: 12x − 3y = 15. Call this (3).

Step 3Add (1) and (3): (2x + 12x) + (3y − 3y) = 13 + 15 → 14x = 28 → x = 2

Step 4Substitute x = 2 into (1): 4 + 3y = 13 → 3y = 9 → y = 3

CheckEquation (2): 4(2) − 3 = 8 − 3 = 5 ✓

x = 2, y = 3

Example 5 — SubstitutionSolve: y = 3x − 1 and 2x + y = 9

Step 1Equation (1) already expresses y in terms of x: y = 3x − 1

Step 2Substitute into (2): 2x + (3x − 1) = 9 → 5x − 1 = 9 → 5x = 10 → x = 2

Step 3Find y: y = 3(2) − 1 = 5

x = 2, y = 5

Example 6 — Three variablesSolve: x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2

Step 1Label (1), (2), (3). Add (1) and (2): 3x + 2z = 9 — call this (4)

Step 2Add (1) and (3): 2x + 3y = 8 — call this (5). Wait — eliminate z instead.

Step 3From (1) − (3): −y + 2z = 4, so y = 2z − 4 — call this (5)

Step 4Substitute (5) into (1): x + (2z − 4) + z = 6 → x + 3z = 10 — call this (6)

Step 5From (4) and (6): 3x + 2z = 9 and x + 3z = 10. Solve: x = 1, z = 3, then y = 2(3)−4 = 2

x = 1, y = 2, z = 3

Core Secondary

Quadratic Equations

Secondary

A quadratic equation has the standard form ax² + bx + c = 0 where a ≠ 0. The word "quadratic" comes from the Latin quadratus (square). Quadratics model an enormous range of real-world situations — projectile motion, profit maximisation, areas, and many physics and engineering problems. The CBC curriculum requires fluency in all three solution methods: factorisation, completing the square, and the quadratic formula.

Standard Form & Quadratic Formula

ax² + bx + c = 0 (a ≠ 0)

x = −b ± √(b² − 4ac) / 2a

The Discriminant — Δ = b² − 4ac

Before solving, compute the discriminant to predict the nature of the roots. This is a favourite short-answer question in KCSE Paper 1.

Δ > 0

Two distinct real roots

e.g. x² − 5x + 6 = 0 → x = 2, 3

Δ = 0

One repeated real root

e.g. x² − 4x + 4 = 0 → x = 2 (twice)

Δ < 0

No real roots (complex)

e.g. x² + x + 1 = 0 → complex roots

Method 1 — Factorisation

Example 1Solve: x² + 5x + 6 = 0 by factorisation

Step 1Ensure the equation equals zero (it does)

Step 2Find two numbers that multiply to c = 6 and add to b = 5. Those are 2 and 3 (2 × 3 = 6; 2 + 3 = 5)

Step 3Write as product of factors: (x + 2)(x + 3) = 0

Step 4Zero Product Property: x + 2 = 0 → x = −2, or x + 3 = 0 → x = −3

Check(−2)² + 5(−2) + 6 = 4 − 10 + 6 = 0 ✓

x = −2 or x = −3

Example 2Factorise: 6x² + 11x + 4

Step 1Multiply a × c = 6 × 4 = 24. Find two numbers multiplying to 24 and adding to 11: 3 and 8.

Step 2Split the middle term: 6x² + 3x + 8x + 4

Step 3Group and factor: 3x(2x + 1) + 4(2x + 1) = (3x + 4)(2x + 1)

(3x + 4)(2x + 1)

Method 2 — Quadratic Formula

Example 3Solve: 2x² − 4x − 6 = 0

Step 1Identify: a = 2, b = −4, c = −6

Step 2Compute discriminant: Δ = (−4)² − 4(2)(−6) = 16 + 48 = 64

Step 3Since Δ = 64 > 0, there are two distinct real roots

Step 4Apply formula: x = (4 ± √64) / 4 = (4 ± 8) / 4

Step 5x = (4 + 8)/4 = 12/4 = 3 or x = (4 − 8)/4 = −4/4 = −1

Check2(3)² − 4(3) − 6 = 18 − 12 − 6 = 0 ✓

x = 3 or x = −1

Method 3 — Completing the Square

Example 4Solve: x² + 6x + 5 = 0 by completing the square

Step 1Move the constant: x² + 6x = −5

Step 2Take half the coefficient of x, then square it: (6/2)² = 9

Step 3Add 9 to both sides: x² + 6x + 9 = −5 + 9 = 4

Step 4Write left side as a perfect square: (x + 3)² = 4

Step 5Take square roots: x + 3 = ±2

Step 6x = −3 + 2 = −1 or x = −3 − 2 = −5

x = −1 or x = −5

Example 5 — Vertex FormWrite y = x² + 4x − 3 in vertex form and state the vertex.

Step 1Take half the coefficient of x: (4/2)² = 4

Step 2Add and subtract 4 inside: y = (x² + 4x + 4) − 4 − 3

Step 3y = (x + 2)² − 7

y = (x + 2)² − 7 Vertex: (−2, −7)

Vieta's Formulas — Sum & Product of Roots

If α and β are the roots of ax² + bx + c = 0, then:

Sum of rootsα + β = −b/a

Product of rootsαβ = c/a

Equation from rootsx² − (α + β)x + αβ = 0

Example 6 — Vieta'sThe roots of x² + px + 12 = 0 are α and 3. Find p and α.

Step 1Product of roots: α × 3 = 12 (from c/a = 12/1) → α = 4

Step 2Sum of roots: α + 3 = −p → 4 + 3 = −p → p = −7

α = 4, p = −7

Core Secondary

Polynomials & Factor Theorem

Secondary

A polynomial in x is an expression of the form aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀ where aₙ ≠ 0 and all exponents are non-negative integers. The degree of the polynomial is n. The Remainder Theorem and Factor Theorem are powerful tools for factorising higher-degree polynomials without long division.

Remainder Theorem & Factor Theorem

Remainder TheoremWhen p(x) ÷ (x − a), remainder = p(a)

Factor Theorem(x − a) is a factor of p(x) ⟺ p(a) = 0

Worked Examples — Polynomials

Example 1Find the remainder when p(x) = 2x³ − 3x² + x − 5 is divided by (x − 2).

Step 1By Remainder Theorem, remainder = p(2)

Step 2p(2) = 2(8) − 3(4) + 2 − 5 = 16 − 12 + 2 − 5 = 1

Remainder = 1

Example 2Show that (x − 2) is a factor of p(x) = x³ − 3x² + x + 2, then fully factorise p(x).

Step 1Compute p(2): p(2) = 8 − 12 + 2 + 2 = 0. Since p(2) = 0, (x − 2) is a factor. ✓

Step 2Divide p(x) by (x − 2) using long division or synthetic division to get the quotient: x² − x − 1

Step 3So p(x) = (x − 2)(x² − x − 1). Check if x² − x − 1 factors: Δ = 1 + 4 = 5, not a perfect square, so it cannot be factorised over integers.

p(x) = (x − 2)(x² − x − 1)

Example 3Fully factorise: p(x) = x³ + 2x² − 5x − 6

Step 1Try integer factors of −6: ±1, ±2, ±3, ±6. Test p(1): 1 + 2 − 5 − 6 = −8 ≠ 0

Step 2Test p(−1): −1 + 2 + 5 − 6 = 0 ✓ So (x + 1) is a factor

Step 3Divide: p(x) ÷ (x + 1) = x² + x − 6

Step 4Factorise x² + x − 6: find numbers multiplying to −6, adding to 1: they are 3 and −2. So (x + 3)(x − 2)

p(x) = (x + 1)(x + 3)(x − 2)

Core Secondary

Inequalities & Functions

Secondary

Critical Rule: When you multiply or divide both sides of an inequality by a negative number, you must reverse (flip) the inequality sign. This is the most common source of errors in KCSE inequality questions.

Linear Inequalities

Example 1Solve: −2x + 5 > 11

Step 1Subtract 5 from both sides: −2x > 6

Step 2Divide by −2 and flip the sign: x < −3

x < −3

Example 2Solve: 3 ≤ 2x + 1 < 9 (compound inequality)

Step 1Subtract 1 from all three parts: 2 ≤ 2x < 8

Step 2Divide all three parts by 2: 1 ≤ x < 4

1 ≤ x < 4

Quadratic Inequalities

To solve a quadratic inequality, find the roots first (treating it as an equation), then use a sign chart or sketch to determine where the expression is positive or negative.

Example 3Solve: x² − 4x − 5 > 0

Step 1Solve x² − 4x − 5 = 0: (x − 5)(x + 1) = 0 → x = 5 or x = −1

Step 2The parabola opens upward (a = 1 > 0), so the expression is positive outside the roots

Step 3The expression is positive when x < −1 or x > 5

x < −1 or x > 5

Example 4Solve: x² − 2x − 3 ≤ 0

Step 1Solve x² − 2x − 3 = 0: (x − 3)(x + 1) = 0 → x = 3 or x = −1

Step 2Parabola opens upward, so expression is ≤ 0 between the roots (inclusive)

−1 ≤ x ≤ 3

Functions — Domain, Range, Composite & Inverse

A function f: A → B assigns exactly one output in B to every input in A. The domain is the set of valid inputs; the range is the set of all outputs.

Composite function(f ∘ g)(x) = f(g(x))

Inverse functionf⁻¹(x): swap x and y, then solve for y

Key propertyf(f⁻¹(x)) = x = f⁻¹(f(x))

Example 5If f(x) = 2x + 3 and g(x) = x², find (f ∘ g)(x) and (g ∘ f)(x).

Step 1(f ∘ g)(x) = f(g(x)) = f(x²) = 2(x²) + 3 = 2x² + 3

Step 2(g ∘ f)(x) = g(f(x)) = g(2x + 3) = (2x + 3)²

(f ∘ g)(x) = 2x² + 3 ; (g ∘ f)(x) = (2x + 3)²

Example 6Find the inverse of f(x) = (2x + 3) / 5

Step 1Replace f(x) with y: y = (2x + 3) / 5

Step 2Swap x and y: x = (2y + 3) / 5

Step 3Solve for y: 5x = 2y + 3 → 2y = 5x − 3 → y = (5x − 3) / 2

f⁻¹(x) = (5x − 3) / 2

Core Secondary

Logarithms & Exponential Functions

Secondary

The logarithm is the inverse of the exponential function. If aˣ = y, then log_a(y) = x. Logarithms were originally developed by John Napier in the 17th century to simplify multiplication by converting it to addition — a remarkable insight. In the CBC curriculum, logarithms appear in Pure Mathematics and are tested in KCSE Paper 1 under the Algebra strand. Their applications extend to calculus, statistics (probability distributions), and real-world problem solving (compound interest, population growth, pH scale).

Definition & Key Relationship

log_a(y) = x ⟺ aˣ = y (a > 0, a ≠ 1, y > 0)

Laws of Logarithms

Product rulelog(ab) = log a + log b

Quotient rulelog(a/b) = log a − log b

Power rulelog(aⁿ) = n log a

Change of baselog_b(x) = ln x / ln b = log x / log b

Special valueslog_a 1 = 0 log_a a = 1

Inverse identitiesa^(log_a x) = x log_a(aˣ) = x

Worked Examples — Logarithms

Example 1Simplify: log 20 + log 5 − log 2

Step 1Apply product rule to the first two: log 20 + log 5 = log(20 × 5) = log 100

Step 2Apply quotient rule: log 100 − log 2 = log(100/2) = log 50

log 50

Example 2Solve: 3^(x + 1) = 27

Step 1Express 27 as a power of 3: 27 = 3³

Step 2Since bases are equal: x + 1 = 3 → x = 2

x = 2

Example 3Solve: 2^(x+1) = 5 (give answer to 3 d.p.)

Step 1Take log (base 10) of both sides: (x + 1) log 2 = log 5

Step 2x + 1 = log 5 / log 2 = 0.69897 / 0.30103 ≈ 2.3219

Step 3x = 2.3219 − 1 = 1.322 (3 d.p.)

x ≈ 1.322

Example 4Solve: log₂(3x − 1) = 4

Step 1Convert from log form to exponential form: 3x − 1 = 2⁴ = 16

Step 23x = 17 → x = 17/3

x = 17/3

Example 5 — Real WorldA population grows according to P = 1000 × 2^(t/5). After how many years does the population reach 8000?

Step 1Set up: 8000 = 1000 × 2^(t/5) → 8 = 2^(t/5)

Step 2Express 8 = 2³: 2^(t/5) = 2³ → t/5 = 3 → t = 15

After 15 years

Advanced

Binomial Theorem

Advanced

The Binomial Theorem provides a formula for expanding (a + b)ⁿ for any positive integer n. Rather than multiplying out the brackets n times, the theorem gives each term directly using binomial coefficients. These coefficients form Pascal's Triangle and are computed using combinations.

Binomial Theorem

(a + b)ⁿ = Σ C(n,r) aⁿ⁻ʳ bʳ where C(n,r) = n! / [r!(n−r)!]

General term: T_(r+1) = C(n,r) · aⁿ⁻ʳ · bʳ

Worked Examples — Binomial Theorem

Example 1Expand (2x + 3)⁴

Step 1Identify a = 2x, b = 3, n = 4. Compute C(4,0), C(4,1), C(4,2), C(4,3), C(4,4) = 1, 4, 6, 4, 1

Step 2T₁ = 1·(2x)⁴·3⁰ = 16x⁴

Step 3T₂ = 4·(2x)³·3¹ = 4·8x³·3 = 96x³

Step 4T₃ = 6·(2x)²·3² = 6·4x²·9 = 216x²

Step 5T₄ = 4·(2x)¹·3³ = 4·2x·27 = 216x; T₅ = 1·1·81 = 81

16x⁴ + 96x³ + 216x² + 216x + 81

Example 2Find the term independent of x in the expansion of (x + 2/x)⁶

Step 1General term: T_(r+1) = C(6,r) · x^(6−r) · (2/x)^r = C(6,r) · 2^r · x^(6−r−r) = C(6,r) · 2^r · x^(6−2r)

Step 2For the term to be independent of x, the power of x must be zero: 6 − 2r = 0 → r = 3

Step 3T₄ = C(6,3) · 2³ = 20 · 8 = 160

The term independent of x is 160

Advanced

Partial Fractions

Advanced

Partial fraction decomposition reverses the process of adding algebraic fractions. It expresses a single rational fraction as a sum of simpler fractions whose denominators are the factors of the original denominator. This technique is essential for integration in calculus and is a Senior Secondary / A-Level topic.

Partial Fraction Forms

Distinct linear factorsp(x)/[(x−a)(x−b)] = A/(x−a) + B/(x−b)

Repeated linear factorp(x)/[(x−a)(x−b)²] = A/(x−a) + B/(x−b) + C/(x−b)²

Irreducible quadraticp(x)/[(x−a)(x²+b)] = A/(x−a) + (Bx+C)/(x²+b)

Worked Examples — Partial Fractions

Example 1Decompose: 5x / [(x−1)(x+2)]

Step 1Write: 5x / [(x−1)(x+2)] = A/(x−1) + B/(x+2)

Step 2Multiply both sides by (x−1)(x+2): 5x = A(x+2) + B(x−1)

Step 3Let x = 1: 5 = 3A → A = 5/3

Step 4Let x = −2: −10 = −3B → B = 10/3

(5/3)/(x−1) + (10/3)/(x+2)

Example 2Decompose: (3x + 5) / [(x+1)(x−2)]

Step 1Write: (3x + 5) / [(x+1)(x−2)] = A/(x+1) + B/(x−2)

Step 23x + 5 = A(x−2) + B(x+1)

Step 3Let x = −1: 2 = −3A → A = −2/3. Let x = 2: 11 = 3B → B = 11/3

(−2/3)/(x+1) + (11/3)/(x−2)

Quick Reference

Algebra Formula Sheet

All key formulae from this lesson in one place. Print or bookmark this section for revision.

Indices

aᵐ · aⁿ = aᵐ⁺ⁿ

aᵐ / aⁿ = aᵐ⁻ⁿ

(aᵐ)ⁿ = aᵐⁿ

a^(m/n) = (ⁿ√a)ᵐ

a⁰ = 1, a⁻ⁿ = 1/aⁿ

Identities

(a+b)² = a²+2ab+b²

(a−b)² = a²−2ab+b²

(a+b)(a−b) = a²−b²

a³+b³ = (a+b)(a²−ab+b²)

a³−b³ = (a−b)(a²+ab+b²)

Quadratics

x = (−b ± √(b²−4ac)) / 2a

Δ = b²−4ac

α + β = −b/a

αβ = c/a

Logarithms

log(ab) = log a + log b

log(a/b) = log a − log b

log(aⁿ) = n log a

log_b(x) = ln x / ln b

Quick Reference

Common Errors to Avoid

These are the most frequent algebra mistakes in KCSE examinations. Being aware of them is the first step to avoiding them.

✗ Wrong (a + b)² = a² + b²

✓ Correct (a + b)² = a² + 2ab + b²

The middle term 2ab is always forgotten.

✗ Wrong −(a − b) = −a − b

✓ Correct −(a − b) = −a + b

A negative outside a bracket changes every sign inside.

✗ Wrong Flipping inequality when dividing by positive

✓ Correct Only flip when dividing/multiplying by a negative

Many students flip the sign regardless of the sign of the divisor.

✗ Wrong log(a + b) = log a + log b

✓ Correct log(ab) = log a + log b

The product rule applies to log of a product, not a sum.

✗ Wrong Writing only one root for a quadratic

✓ Correct Always check if ± gives two roots

The ± sign means there are (usually) two solutions.

✗ Wrong Solving x² = 9 as x = 3 only

✓ Correct x = ±3 (both 3 and −3 satisfy it)

Taking the square root always gives ± unless stated otherwise.