Mathematics Grade 12 · Year Summary

Mathematics · Grade 12 · Year Summary Date: Top Journal

Every distinction begins with one topic at a time.- start now

1 · Algebra & Equations
🔒 2 · Sequences & Series
🔒 3 · Functions & Inverses
🔒 4 · Logarithms
🔒 5 · Finance, Growth & Decay
🔒 6 · Polynomials
🔒 7 · Differential Calculus
🔒 8 · Probability
🔒 9 · Analytical Geometry
🔒 10 · Compound Angles & 3D Trig
🔒 11 · Trigonometric Graphs
🔒 12 · Euclidean Geometry
🔒 13 · Statistics & Regression
🔒 14 · Reference Sheet
🔒 Journal
🔒 Wellbeing
🔒 Planner

Paper 1 · Algebra and Equations 25 marks

CHAPTER 1 · THE FOUNDATION OF PAPER 1 ALGEBRA & EQUATIONS

Learning outcomes · you must be able to

Factorise fluently: common factor, difference of squares, trinomials, grouping, and the sum and difference of two cubes.
Solve a quadratic equation by factorising or by applying the quadratic formula, and round correctly when asked.
Use the discriminant Δ = b² − 4ac to describe the nature of the roots, and work backwards to find the value of an unknown constant.
Solve a quadratic inequality and write the answer in the correct interval form.
Solve two equations simultaneously when one is linear and the other is quadratic.
Solve exponential equations, including those hiding a quadratic behind a substitution.
Solve equations containing a square root, and test every answer back in the original equation.

📖 KEY DEFINITION A quadratic equation is any equation that can be written as ax² + bx + c = 0 with a ≠ 0. Its roots (also called solutions or zeros) are the x-values that make the statement true. The discriminant is the expression Δ = b² − 4ac sitting underneath the square root in the formula; on its own it tells you what KIND of roots the equation has without ever solving it. Rational roots are numbers that can be written as a fraction of two integers; irrational roots cannot, and they always arrive with a surd still attached.

Core content · factorising, the skill everything else stands on

Almost every mark in this section, and a surprising number of marks in Functions, Calculus and Number Patterns, comes back to whether you can factorise quickly and correctly. The order never changes: take out a common factor first, then look at how many terms are left.

What you see	Pattern	Own example
Every term shares something	Common factor first, always	6x³ − 15x² = 3x²(2x − 5)
Two terms, both squares, a minus between them	a² − b² = (a − b)(a + b)	49x² − 25 = (7x − 5)(7x + 5)
Three terms, x² leading	Trinomial: two numbers that multiply to ac and add to b	x² − 2x − 15 = (x − 5)(x + 3)
Four terms	Group in pairs, then factor the bracket out	x³ + 2x² − 9x − 18 = (x + 2)(x² − 9)
Two terms, both cubes	a³ ± b³ = (a ± b)(a² ∓ ab + b²)	8x³ − 27 = (2x − 3)(4x² + 6x + 9)

⚡ MUST MEMORISE The cube pattern is the one learners forget, and it is not on the information sheet. Remember it as SOAP: Same sign in the first bracket, Opposite sign in the middle of the second bracket, Always Positive at the end of the second bracket. So a³ − b³ = (a − b)(a² + ab + b²) and a³ + b³ = (a + b)(a² − ab + b²).

Worked example 1 · factorise fully, in the right order

Q: Factorise 2x³ − 32x completely.

Common factor first: 2x(x² − 16)

The bracket is a difference of squares: x² − 16 = (x − 4)(x + 4)

∴ 2x³ − 32x = 2x(x − 4)(x + 4)

Note what going straight for the difference of squares would have cost you: without pulling out the 2x first, the expression is not a difference of squares at all, and the mark is gone.

Worked example 2 · factorising by grouping

Q: Factorise 3x³ − 12x² − 4x + 16.

Group in pairs: (3x³ − 12x²) + (−4x + 16)

Take a factor out of each pair: 3x²(x − 4) − 4(x − 4)

The bracket (x − 4) is now common: (x − 4)(3x² − 4)

∴ 3x³ − 12x² − 4x + 16 = (x − 4)(3x² − 4)

The sign check: pulling −4 out of −4x + 16 leaves (x − 4), matching the first bracket. If the two brackets do not match, your second factor has the wrong sign.

Core content · solving a quadratic equation

There are only two routes the examiner expects. Try factorising first, because it is faster and cleaner. If the trinomial refuses to factorise, or the question asks for an answer correct to two decimal places, that is the paper telling you to use the formula.

⚡ MUST MEMORISE

x = [−b ± √(b² − 4ac)] ÷ 2a is printed on the information sheet, so you do not need to memorise the symbols. What you DO need automatically is the setup: the equation must be written in the form ax² + bx + c = 0, with everything on one side and 0 on the other, BEFORE you read off a, b and c.

Factorising route

Write the equation as … = 0.
Factorise the left-hand side.
Set each factor equal to 0 separately.
Solve each small equation.

Formula route

Write the equation as ax² + bx + c = 0.
Write down a, b and c, with their signs.
Substitute into the formula and show the substitution line.
Give both roots, rounded only at the very last step.

Worked example 3 · solving by factorising

Q: Solve for x: 2x² + 5x = 12.

Move everything to one side: 2x² + 5x − 12 = 0

ac = 2 × (−12) = −24, and the pair that multiplies to −24 and adds to 5 is 8 and −3

2x² + 8x − 3x − 12 = 0, so 2x(x + 4) − 3(x + 4) = 0

(x + 4)(2x − 3) = 0

∴ x = −4 or x = 1,5

Worked example 4 · solving with the formula, rounded

Q: Solve for x, correct to TWO decimal places: 3x² − 7x − 1 = 0.

a = 3, b = −7, c = −1

Δ = (−7)² − 4(3)(−1) = 49 + 12 = 61

x = [7 ± √61] ÷ 6

x = (7 + 7,8102…) ÷ 6 or x = (7 − 7,8102…) ÷ 6

∴ x = 2,47 or x = −0,14

Notice that −b became +7 because b itself was negative. That single sign is the most common formula error in the whole paper.

⚠️ COMMON MISTAKE Completing the square is not an examinable method for solving a quadratic equation, so building your whole strategy around it wastes time you do not have. You still meet the technique when a parabola has to be written in turning-point form y = a(x + p)² + q, which is Functions work, not equation-solving. In this section: factorise, or use the formula.

⚠️ COMMON MISTAKE Dividing both sides by x to "simplify" x² = 4x. That throws away the root x = 0 and costs you half the answer every time. Move everything to one side instead: x² − 4x = 0, so x(x − 4) = 0, giving x = 0 or x = 4. You may only divide by something you can prove is never zero.

Core content · the nature of the roots

The discriminant is a description question dressed as a calculation. You are not asked for the roots, you are asked what sort of roots exist, and the sign of Δ answers it completely.

Value of Δ = b² − 4ac	Nature of the roots	What the parabola does
Δ < 0	Non-real (no real roots)	Never touches the x-axis
Δ = 0	Real, equal, rational	Touches the x-axis exactly once
Δ > 0, a perfect square	Real, unequal, rational	Cuts the x-axis at two "neat" points
Δ > 0, not a perfect square	Real, unequal, irrational	Cuts the x-axis at two surd points

⭐ EXAM FAVOURITE "For which value(s) of k will the equation have equal roots / two real roots / no real roots?" This question appears in some form in almost every sitting. The method never varies: write the equation in standard form, express Δ in terms of k, then convert the description into an inequality or equation on Δ and solve it. Half the marks are earned before you touch k at all, simply by writing Δ = b² − 4ac correctly with the k-terms in place.

Worked example 5 · describing the nature of the roots

Q: Without solving the equation, describe the nature of the roots of 2x² − 3x − 5 = 0.

a = 2, b = −3, c = −5

Δ = (−3)² − 4(2)(−5) = 9 + 40 = 49

49 > 0 and 49 = 7², so it is a perfect square

∴ the roots are real, unequal and rational

Worked example 6 · working backwards to find k

Q: Determine the value(s) of k for which x² + (k + 2)x + 9 = 0 has equal roots.

a = 1, b = k + 2, c = 9

Equal roots means Δ = 0

(k + 2)² − 4(1)(9) = 0

k² + 4k + 4 − 36 = 0, so k² + 4k − 32 = 0

(k + 8)(k − 4) = 0

∴ k = −8 or k = 4

The discriminant question turned into an ordinary quadratic, just in k instead of x. That is the pattern every time.

Worked example 7 · a k-question that ends in an inequality

Q: For which values of p will x² − 6x + p = 0 have two unequal real roots?

a = 1, b = −6, c = p

Two unequal real roots means Δ > 0

36 − 4p > 0

−4p > −36

∴ p < 9

Dividing an inequality by a negative number flips the sign. Skipping that flip turns a correct method into a wrong answer.

Read the wording of every k-question twice. "Real roots" allows Δ ≥ 0, but "two unequal real roots" demands Δ > 0 strictly, and "non-real" needs Δ < 0. One symbol decides the whole answer.

Core content · quadratic inequalities

An inequality is solved in three moves: get zero on one side, find the critical values by pretending it is an equation, then decide which region satisfies the sign. A quick rough parabola, drawn in the margin, decides that last step faster than any table.

If the parabola opens upwards

< 0 means the graph is BELOW the x-axis, which is the piece BETWEEN the roots.
> 0 means the graph is ABOVE the x-axis, which is the two pieces OUTSIDE the roots.

Writing the answer

Between the roots: one statement, x₁ < x < x₂.
Outside the roots: two statements joined by "or", never by "and".
Use ≤ and ≥ only if the original inequality was ≤ or ≥.

Worked example 8 · an answer that lies BETWEEN the roots

Q: Solve for x: x² − x − 6 < 0.

Critical values: x² − x − 6 = 0, so (x − 3)(x + 2) = 0, giving x = 3 and x = −2

a = 1 > 0, so the parabola opens upwards and dips below the axis between the roots

We want the part that is below the axis

∴ −2 < x < 3

Worked example 9 · an answer that lies OUTSIDE the roots

Q: Solve for x: 2x² + 3x ≥ 2.

Zero on one side: 2x² + 3x − 2 ≥ 0

Critical values: (2x − 1)(x + 2) = 0, so x = 0,5 and x = −2

a = 2 > 0, and we want the parts at or above the axis, which sit outside the roots

∴ x ≤ −2 or x ≥ 0,5

⚠️ COMMON MISTAKE Writing an "outside" answer as a single chain, for example 0,5 ≤ x ≤ −2. That statement is impossible, because no number is at the same time bigger than 0,5 and smaller than −2. An outside answer is always two separate statements joined by the word "or". Markers treat the chained version as wrong even when the critical values are right.

💡 EASY MARK After solving any inequality, test one number from your answer region in the ORIGINAL inequality. In worked example 8, x = 0 gives 0 − 0 − 6 = −6, which is indeed less than 0, so the region is right. Ten seconds of testing protects three or four marks.

Core content · simultaneous equations, one linear and one quadratic

The examiner always gives you one straight-line equation and one curved one. Make a variable the subject of the LINEAR equation, because that is the one with no squares in it, then substitute into the quadratic. Two pairs of answers come out, and both must be stated as pairs.

Worked example 10 · substitution done in the right order

Q: Solve for x and y simultaneously: y − x = 2 and x² + y² = 100.

From the linear equation: y = x + 2

Substitute: x² + (x + 2)² = 100

x² + x² + 4x + 4 = 100, so 2x² + 4x − 96 = 0

Divide through by 2: x² + 2x − 48 = 0, so (x + 8)(x − 6) = 0

x = −8 or x = 6

Back into y = x + 2: if x = −8 then y = −6; if x = 6 then y = 8

∴ (x ; y) = (−8 ; −6) or (6 ; 8)

⚠️ COMMON MISTAKE Solving for x and then stopping, or pairing the answers wrongly: writing x = −8 with y = 8. Always substitute each x back into the LINEAR equation, never into the quadratic, and keep the pairs together on the same line. Half a question's marks live in the pairing.

Core content · exponential equations

When the unknown sits in the exponent, you have exactly two tools: make the bases equal so the exponents can be equated, or spot a hidden quadratic and substitute a single letter for the repeated power.

Worked example 11 · making the bases equal

Q: Solve for x: 3^2x−1 = 81.

Write 81 as a power of 3: 81 = 3⁴

3^2x−1 = 3⁴

Equal bases, so equate the exponents: 2x − 1 = 4

∴ x = 2,5

Worked example 12 · a quadratic hiding inside an exponential

Q: Solve for x: 2^2x − 6·2^x + 8 = 0.

Note that 2^2x = (2^x)², so let k = 2^x

k² − 6k + 8 = 0, so (k − 2)(k − 4) = 0

k = 2 or k = 4

Replace k: 2^x = 2 gives x = 1; and 2^x = 4 = 2² gives x = 2

∴ x = 1 or x = 2

🔥 FREQUENTLY TESTED A substitution question that produces a NEGATIVE value for k. Since 2^x (or any a^x with a > 0) can never be negative or zero, that value must be rejected in writing. Say it out loud on the page: "2^x = −3 has no solution, since 2^x > 0 for all x". The rejection sentence itself carries a mark.

Core content · equations containing a square root

Squaring both sides is legal, but it is not reversible: squaring can manufacture answers that never satisfied the original equation. Those false answers are called extraneous roots, and the only defence is testing.

Worked example 13 · isolate, square, then test

Q: Solve for x: √(x + 7) = x + 1.

The root is already alone on the left, so square both sides

x + 7 = (x + 1)² = x² + 2x + 1

0 = x² + x − 6, so (x + 3)(x − 2) = 0

x = −3 or x = 2

Test x = 2: √9 = 3 and 2 + 1 = 3, so it works

Test x = −3: √4 = 2 but −3 + 1 = −2, so it fails

∴ x = 2 only

⚠️ COMMON MISTAKE Handing in both answers because both came out of a correct quadratic. The mark for a surd equation is awarded for the rejection, not for the algebra. Write the test line for each root, then state clearly which one you discard and why.

Quick reference · exponent laws and surd forms

Law	Statement	Own example
Multiply	a^m × aⁿ = a^m+n	2³ × 2⁴ = 2⁷
Divide	a^m ÷ aⁿ = a^m−n	5⁶ ÷ 5² = 5⁴
Power of a power	(a^m)ⁿ = a^mn	(3²)⁵ = 3¹⁰
Zero exponent	a⁰ = 1 for a ≠ 0	(7x)⁰ = 1
Negative exponent	a⁻ⁿ = 1 ÷ aⁿ	4⁻² = 1⁄16
Fractional exponent	a^m/n = the n-th root of a^m	8^2/3 = 4

💡 EASY MARK Simplification questions that mix bases, such as (12^x × 9^x) ÷ 6^2x, are pure prime-factor work: rewrite 12 as 2²·3, 9 as 3² and 6 as 2·3, then let the laws cancel everything. No new theory is needed, only patience with primes.

Connections to other topics

→ Functions and graphs: the roots of ax² + bx + c = 0 ARE the x-intercepts of the parabola y = ax² + bx + c, and the discriminant tells you in advance how many intercepts to expect.
→ Differential calculus: solving f′(x) = 0 for a cubic is a quadratic equation, so every turning-point mark rests on the factorising you practise here.
→ Analytical geometry: finding where a line cuts a circle is exactly the linear-plus-quadratic simultaneous system in worked example 10.
→ Sequences and series: quadratic patterns produce a general term T_n that is a quadratic in n, and finding which term equals a given value is one more quadratic equation.

Algebra is not one topic on the paper. It is the language the rest of the paper is written in.- one factorisation at a time

🧠 ACTIVE RECALL

The discriminant of ax² + bx + c = 0 is given by:(1)

If Δ = −12 for a quadratic equation, its roots are:(2)

The solution of x² − 4x − 5 < 0 is:(2)

The full factorisation of x³ − 8 is:(2)

If 3^x+2 = 27, then x equals:(2)

Why must every answer to a surd equation be substituted back into the original equation?(2)

Solve for x, correct to TWO decimal places: 2x² − 5x = 4.(4)

Determine the value(s) of k for which x² + kx + 9 = 0 will have equal roots.(5)

Solve for x and y simultaneously: 2x + y = 5 and xy = 3.(6)

Explain to a Grade 8: what does the discriminant tell you about a quadratic equation, WITHOUT solving it?

ONE-MINUTE SUMMARY

Common factor first, then count the terms: two squares means difference of squares, three terms means a trinomial, four terms means grouping, two cubes means SOAP. Solve quadratics by factorising, or by the formula when the question asks for decimals. Δ = b² − 4ac describes the roots: negative means non-real, zero means equal, positive means unequal, and a perfect square makes them rational. Inequalities need critical values, a rough parabola and interval notation, with "or" for outside answers. Simultaneous systems: make a variable the subject of the LINEAR equation, substitute, then pair the answers. Exponentials: equal bases, or substitute k for a^x and reject any negative k. Surd equations: isolate, square, and test both roots back in the original.

IN THE EXAM

Where: Paper 1, Question 1, worth about 25 marks. It is the opening question of the whole examination and it is the most predictable question on either paper.
Typical Qs: "Solve for x" in three or four short parts of rising difficulty, then "determine the value(s) of k for which …", then a simultaneous system, and often a short exponent-law simplification to close.
What earns marks: writing the equation in standard form before quoting a, b and c, showing the substitution line in the formula before the answer line, stating the parabola's direction when solving an inequality, giving simultaneous answers as PAIRS, and writing the rejection sentence in full for a negative k or an extraneous surd root.
Time plan: aim to be finished with this question inside 25 minutes. It is designed to be scored heavily and quickly, and the marks lost here are almost never lost to difficulty, they are lost to signs and to skipped steps.

🔒 SEQUENCES & SERIES

🔒 FUNCTIONS & INVERSES

🔒 LOGARITHMS

🔒 FINANCE, GROWTH & DECAY

🔒 POLYNOMIALS

🔒 DIFFERENTIAL CALCULUS

🔒

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