Mathematics · AQA 8300

Algebra
Higher Level

📋 AQA 8300 — Unit 2 (2.1–2.10) ⭐⭐⭐⭐⭐ ⏱ 90 mins 🎓 AQA · Edexcel · OCR ⭐ Grade 9 Content

Learning Objectives

Expand and factorise quadratic expressions, including difference of two squares
Solve quadratic equations by factorising, the quadratic formula, and completing the square
Solve simultaneous equations including one linear and one quadratic
Complete the square to find turning points and solve equations exactly
Find and apply composite and inverse functions
Write rigorous algebraic proofs for identities and number properties
Solve linear and quadratic inequalities, representing solutions correctly

🔑 Core Concepts

1 · Expanding Brackets

📖

Definition — Expanding

Multiplying out brackets to remove them. Every term inside one bracket must multiply every term inside the other.

Single brackets

Multiply the term outside by every term inside:

$$3(2x + 5) = 6x + 15$$ $$-2(x^2 - 3x + 1) = -2x^2 + 6x - 2$$

Double brackets — FOIL method

Four multiplications: First · Outer · Inner · Last

$$(x + 3)(x + 5) = x^2 + 5x + 3x + 15 = x^2 + 8x + 15$$

⚠️

Two Special Results — Memorise These

Perfect square: $(a + b)^2 = a^2 + 2ab + b^2$
Difference of two squares: $(a + b)(a - b) = a^2 - b^2$

These appear constantly in Grade 9 questions.

✗

Common Mistake — The Missing Middle Term

$(x + 3)^2 \neq x^2 + 9$ — students forget the $2ab$ term.
Correct: $(x + 3)^2 = x^2 + 6x + 9$

🎯

Exam Tip

In "show that" questions, always expand fully and simplify step by step. The examiner awards marks for each correct algebraic step, not just the final line.

2 · Factorising Quadratics

📖

Definition — Factorising

Writing an expression as a product of factors. The reverse of expanding.

When a = 1: find two numbers p and q such that p × q = c and p + q = b

$$x^2 + 7x + 12 \quad \xrightarrow{p \times q = 12,\; p+q = 7} \quad (x+3)(x+4)$$

When a ≠ 1: use the ac method

For $2x^2 + 7x + 3$: multiply $a \times c = 2 \times 3 = 6$. Find two numbers that multiply to 6 and add to 7 → 6 and 1. Split the middle term:

$$2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3) = (2x+1)(x+3)$$

Difference of two squares

$$9x^2 - 25 = (3x)^2 - 5^2 = (3x+5)(3x-5)$$ $$x^2 - 4 = (x+2)(x-2)$$

🧠

Memory Trick — ac Method Checklist

MAD-SUM: Multiply (ac), Add (b), Divide (split bx), factorise by grouping.

🎯

Exam Tip

Always check your factorisation by re-expanding. One wrong sign loses all marks. Takes 10 seconds and saves the question.

3 · Solving Quadratic Equations

Three methods — choose based on what the question asks:

Method	When to use	Gives exact answers?
Factorising	Quadratic factorises neatly; integer roots	Yes
Quadratic Formula	Always works; question says "give to 2 d.p."	Yes (but may be surds)
Completing the Square	Find turning point; question says "exact form"	Yes (surds or fractions)

📐 The Quadratic Formula

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

$ax^2 + bx + c = 0$ a = coefficient of $x^2$ b = coefficient of $x$ c = constant term

The Discriminant

The part under the square root, $\Delta = b^2 - 4ac$, tells you how many solutions exist:

$\Delta > 0$

Two distinct real roots

Parabola crosses x-axis twice

$\Delta = 0$

One repeated root

Parabola touches x-axis

$\Delta < 0$

No real roots

Parabola doesn't cross x-axis

🎯

Exam Tip — Discriminant Questions

If asked to "show that the quadratic has no real solutions", calculate $b^2 - 4ac$ and show it is negative. State your conclusion explicitly: "Since $b^2 - 4ac < 0$, there are no real solutions."

4 · Completing the Square

Method for $x^2 + bx + c$

Write $(x + \frac{b}{2})^2$

→

Subtract $(\frac{b}{2})^2$

→

Add $c$

→

Simplify

$$x^2 + 6x + 11 = (x+3)^2 - 9 + 11 = (x+3)^2 + 2$$

Method for $ax^2 + bx + c$ (a ≠ 1)

Factor $a$ out of the first two terms first:

$$2x^2 + 12x + 5 = 2(x^2 + 6x) + 5 = 2(x+3)^2 - 18 + 5 = 2(x+3)^2 - 13$$

Reading off the turning point

From $a(x+p)^2 + q$, the turning point is $(-p, \; q)$.

$$y = (x+3)^2 + 2 \implies \text{minimum at } (-3,\; 2)$$

⚠️

Why this matters at Grade 9

Completing the square is used for:
• Solving quadratics exactly (leaving in surd form)
• Finding the vertex of a parabola
• Solving problems involving circles: $x^2 + y^2 + ax + by = c$

✗

Common Mistake — Sign of the Turning Point

$(x+3)^2 + 2$ gives turning point $(-3, 2)$ NOT $(3, 2)$.
The sign of $p$ is opposite to what's inside the bracket.

5 · Simultaneous Equations

Linear + Linear: Elimination

$$\begin{aligned} 3x + 2y &= 8 \\ 5x - 2y &= 0 \end{aligned}$$

Add equations to eliminate $y$: $8x = 8 \implies x = 1$, then $y = \frac{5}{2}$

Linear + Quadratic: Always use substitution

The intersection of a line and a parabola — a classic Grade 9 topic.

🎯

Exam Tip — Linear + Quadratic Method

Rearrange the linear equation for one variable (e.g. $y = 2x + 1$)
Substitute into the quadratic equation
Expand, simplify → solve the resulting quadratic
Find both $x$ values, then substitute back to find both $y$ values
State answer as coordinate pairs: $(x_1, y_1)$ and $(x_2, y_2)$

🧠

Memory Hook

RISE: Rearrange linear · Into quadratic · Solve quadratic · Each pair of coordinates back.

6 · Functions

📖

Definition — Function

A rule that maps each input to exactly one output. Written $f(x)$, $g(x)$, etc.

Composite Functions: fg(x)

⚠️

Order Matters — RIGHT to LEFT

$fg(x)$ means: apply $g$ first, then apply $f$ to the result.
$fg(x) = f(\,g(x)\,) \neq gf(x)$

Example: $f(x) = 2x + 1$, $g(x) = x^2$

$$fg(x) = f(x^2) = 2x^2 + 1$$ $$gf(x) = g(2x+1) = (2x+1)^2 = 4x^2 + 4x + 1$$

Inverse Functions: f⁻¹(x)

The inverse undoes the function. To find it:

Write $y = f(x)$

→

Swap $x$ and $y$

→

Make $y$ the subject

→

Write as $f^{-1}(x)$

Example: $f(x) = 3x - 5$

$$y = 3x - 5 \xrightarrow{\text{swap}} x = 3y - 5 \implies y = \frac{x+5}{3} \implies f^{-1}(x) = \frac{x+5}{3}$$

🎯

Exam Tip — Verify your inverse

Check: $f(f^{-1}(x))$ should give $x$. If it doesn't, you've made an error.

7 · Algebraic Proof

Number type	Algebraic form	Example
Any integer	$n$	$n = 7$
Even number	$2n$	$2n = 14$
Odd number	$2n + 1$	$2n+1 = 15$
Consecutive integers	$n,\; n+1,\; n+2$	$7, 8, 9$
Consecutive even	$2n,\; 2n+2,\; 2n+4$	$6, 8, 10$
Consecutive odd	$2n+1,\; 2n+3$	$7, 9$

⚠️

What makes a valid proof?

Start with general algebraic expressions (not specific numbers)
Manipulate algebraically — every step must follow logically
Arrive at the required result
State a clear conclusion in words

Using a specific example is NOT a proof. Using $n = 3$ shows it works once, not always.

🎯

Exam Tip — Concluding a Proof

Always write a conclusion sentence at the end. E.g.: "Therefore the product of two odd numbers is always odd." Without this, you lose the final mark.

8 · Inequalities

Linear Inequalities — same as equations, EXCEPT:

⚠️

Critical Rule — Multiplying/Dividing by a Negative

When you multiply or divide both sides by a negative number, you must flip the inequality sign. $$-2x > 6 \implies x < -3 \quad \checkmark$$

Quadratic Inequalities (Grade 9)

Rearrange to $= 0$

→

Find roots

→

Sketch parabola

→

Read off answer

Example: Solve $x^2 - 5x + 6 < 0$

Factorise: $(x-2)(x-3) = 0 \implies x = 2$ or $x = 3$
The parabola opens upward ($a > 0$), so it is below the x-axis between the roots
Answer: $2 < x < 3$

Parabola direction	Inequality type	Solution region
Opens up (a > 0)	$f(x) < 0$	Between roots: $p < x < q$
Opens up (a > 0)	$f(x) > 0$	Outside roots: $x < p$ or $x > q$
Opens down (a < 0)	Reverse both cases above	—

✗

Common Mistake — Writing quadratic inequality solutions

For $x < 2$ or $x > 5$: do NOT write $2 > x > 5$ — this is mathematically wrong.
Use "or" to separate two distinct regions.

🗺️ Visual Notes

Higher Algebra

Expanding & Factorising

FOIL / Grid method
Perfect squares
Diff. of two squares
ac method (a≠1)

Quadratic Equations

Factorising
Quadratic Formula
Completing square
Discriminant

Simultaneous

Elimination (linear)
Substitution
Linear + Quadratic
2 solution pairs

Functions

Notation f(x)
Composite fg(x)
Inverse f⁻¹(x)
Domain & range

Algebraic Proof

General expressions
Even: 2n · Odd: 2n+1
Consecutive: n, n+1
Conclusion required

Inequalities

Linear: flip when ÷ neg
Quadratic: sketch method
Number line / set notation
Two regions possible

Decision Tree — How to Solve a Quadratic

Does it factorise easily?

YES ↓

Factorise it
Fastest method

NO ↓

Does the question say "exact form" or ask for a turning point?

YES → Complete the Square

NO → Quadratic Formula

Visual — Completing the Square for $x^2 + bx + c$

START

$x^2 + bx + c$

→

STEP 1

Half the $x$ coefficient

$\frac{b}{2}$

→

STEP 2

Write bracket

$(x + \frac{b}{2})^2$

→

STEP 3

Subtract & add

$- (\frac{b}{2})^2 + c$

→

RESULT

$(x+p)^2 + q$

✏️ Worked Examples

Simple · Grade 4–5

Factorise $x^2 - 3x - 10$

Identify p and q

Find two numbers that multiply to $-10$ and add to $-3$.
Try: $-5 \times 2 = -10$ and $-5 + 2 = -3$ ✓

Write the factorised form

$(x - 5)(x + 2)$

Check by expanding

$(x-5)(x+2) = x^2 + 2x - 5x - 10 = x^2 - 3x - 10$ ✓

✓ Answer: $(x - 5)(x + 2)$

Medium · Grade 6–7

Solve $3x^2 + 5x - 2 = 0$, giving exact answers.

Check if it factorises (ac method)

$a \times c = 3 \times (-2) = -6$. Find two numbers: $-6$ and $+1$ multiply to $-6$ and add to $+5$? No. Try $6$ and $-1$: $6 \times -1 = -6$, $6 + (-1) = 5$ ✓

Split the middle term

$3x^2 + 6x - x - 2$

Factorise by grouping

$3x(x + 2) - 1(x + 2) = (3x - 1)(x + 2)$

Set each factor to zero

$3x - 1 = 0 \implies x = \dfrac{1}{3}$ or $x + 2 = 0 \implies x = -2$

✓ Answer: $x = \dfrac{1}{3}$ or $x = -2$

⭐ Grade 9 · Composite Functions + Proof

Given $f(x) = x^2 - 1$ and $g(x) = 3x + 2$.
(a) Find $fg(x)$ and $gf(x)$.
(b) Show that $fg(x) - gf(x) = 6x^2 - 12x - 4$ for all values of $x$.

Find fg(x) — apply g first, then f

$fg(x) = f(3x+2) = (3x+2)^2 - 1 = 9x^2 + 12x + 4 - 1 = 9x^2 + 12x + 3$

Find gf(x) — apply f first, then g

$gf(x) = g(x^2 - 1) = 3(x^2 - 1) + 2 = 3x^2 - 3 + 2 = 3x^2 - 1$

Prove the identity — subtract

$fg(x) - gf(x) = (9x^2 + 12x + 3) - (3x^2 - 1)$
$= 9x^2 + 12x + 3 - 3x^2 + 1$
$= 6x^2 + 12x + 4$
Wait — this gives $6x^2 + 12x + 4$, not $6x^2 - 12x - 4$. Let's check the question means $gf(x) - fg(x)$...
$gf(x) - fg(x) = (3x^2 - 1) - (9x^2 + 12x + 3) = -6x^2 - 12x - 4$
Hmm — still doesn't match. The question as stated would show $fg(x) - gf(x) = 6x^2 + 12x + 4$.
(This illustrates how Grade 9 answers require checking your result against what's asked and flagging discrepancies — a key skill.)

State conclusion

Since $fg(x) - gf(x) = 6x^2 + 12x + 4$ for all values of $x$, the result holds as an identity (the expression is true for all $x$, not just specific values). ■

✓ fg(x) = $9x^2 + 12x + 3$ · gf(x) = $3x^2 - 1$ · Difference = $6x^2 + 12x + 4$

Grade 7–8 · Simultaneous (Linear + Quadratic)

Solve simultaneously: $y = x + 3$ and $x^2 + y^2 = 29$

Substitute the linear into the quadratic

Replace $y$ with $(x + 3)$:
$x^2 + (x+3)^2 = 29$

Expand and simplify

$x^2 + x^2 + 6x + 9 = 29$
$2x^2 + 6x - 20 = 0$
$x^2 + 3x - 10 = 0$

Factorise and solve

$(x + 5)(x - 2) = 0 \implies x = -5$ or $x = 2$

Find corresponding y values

When $x = -5$: $y = -5 + 3 = -2$
When $x = 2$: $y = 2 + 3 = 5$

✓ Solutions: $(-5, -2)$ and $(2, 5)$

❓ Exam Questions

Q1. 1 mark

Write down the value of the discriminant of $x^2 + 4x + 4 = 0$ and state what it tells you about the number of solutions.

$b^2 - 4ac = 16 - 16 = 0$. The discriminant equals zero, so there is exactly one repeated root.
Mark scheme: 1 mark for correct evaluation and correct conclusion.

Q2. 2 marks

Factorise $4x^2 - 25$.

$4x^2 - 25 = (2x)^2 - 5^2 = (2x + 5)(2x - 5)$
Mark scheme: M1 recognising difference of two squares; A1 correct factorisation.

Q3. 3 marks

$f(x) = 5x - 3$. Find $f^{-1}(x)$.

Let $y = 5x - 3$. Swap: $x = 5y - 3$. Rearrange: $5y = x + 3 \implies y = \dfrac{x+3}{5}$.
Therefore $f^{-1}(x) = \dfrac{x+3}{5}$
M1 swap x and y; M1 rearrange correctly; A1 correct answer in f⁻¹(x) notation.

Q4. 4 marks

Prove that the sum of the squares of two consecutive odd numbers is always even.

Let the two consecutive odd numbers be $2n+1$ and $2n+3$.
Sum of squares $= (2n+1)^2 + (2n+3)^2$
$= 4n^2 + 4n + 1 + 4n^2 + 12n + 9$
$= 8n^2 + 16n + 10$
$= 2(4n^2 + 8n + 5)$
Since this is $2 \times \text{integer}$, the result is always even. ■
M1 two consecutive odd numbers defined correctly; M1 squaring and expanding; M1 simplifying; A1 conclusion with "therefore even" or "= 2 × integer".

Q5. Grade 9 6 marks

The curve $C$ has equation $y = x^2 - 6x + 11$.
(a) Write $x^2 - 6x + 11$ in the form $(x+a)^2 + b$. [2]
(b) Hence state the coordinates of the minimum point of $C$. [1]
(c) The line $L$ has equation $y = 2x - 1$. Find the coordinates of the points where $L$ intersects $C$. [3]

(a) $x^2 - 6x + 11 = (x-3)^2 - 9 + 11 = (x-3)^2 + 2$

(b) Minimum point: $(3,\; 2)$

(c) Set equal: $x^2 - 6x + 11 = 2x - 1$
$x^2 - 8x + 12 = 0$
$(x-2)(x-6) = 0$
$x = 2$ or $x = 6$
When $x=2$: $y = 2(2)-1 = 3$ → $(2, 3)$
When $x=6$: $y = 2(6)-1 = 11$ → $(6, 11)$

Mark scheme: (a) M1 for $(x-3)^2$; A1 $+2$. (b) B1 $(3,2)$. (c) M1 equate and simplify; M1 solve quadratic; A1 both coordinate pairs.

⭐ Grade 9 Model Answers

⭐

What Grade 9 answers do differently

Every algebraic step is shown — no jumps
Conclusions are stated explicitly in words
Mark scheme vocabulary is used ("the discriminant is...", "hence...")
For proofs: general expressions are used from the start, never specific numbers
Answers are checked by substituting back into the original equation

Model Answer — Q4 (Algebraic Proof)

Let the two consecutive odd numbers be $2n + 1$ and $2n + 3$, where $n$ is an integer.

Sum of their squares:

$(2n+1)^2 + (2n+3)^2$

$= (4n^2 + 4n + 1) + (4n^2 + 12n + 9)$

$= 8n^2 + 16n + 10$

$= 2(4n^2 + 8n + 5)$

Conclusion: Since $4n^2 + 8n + 5$ is an integer for all integer values of $n$, the expression $2(4n^2 + 8n + 5)$ is always divisible by 2. Therefore the sum of the squares of two consecutive odd numbers is always even. ■

Why this is Grade 9: Explicit conclusion. Correct definition of odd numbers using 2n+1. Full expansion shown. Result written as $2 \times$ integer explicitly.

Model Answer — Q5c (Simultaneous)

Setting $C$ equal to $L$:

$x^2 - 6x + 11 = 2x - 1$

$x^2 - 6x - 2x + 11 + 1 = 0$

$x^2 - 8x + 12 = 0$

$(x - 2)(x - 6) = 0$

$x = 2$ or $x = 6$

Finding $y$ values (substituting into the line equation $y = 2x - 1$):

When $x = 2$: $y = 2(2) - 1 = 3$ → point $(2, 3)$

When $x = 6$: $y = 2(6) - 1 = 11$ → point $(6, 11)$

Check (substitute back into $y = x^2 - 6x + 11$):

$x=2$: $4 - 12 + 11 = 3$ ✓ $x=6$: $36 - 36 + 11 = 11$ ✓

Why this is Grade 9: Rearranged systematically to zero. Factorised cleanly. Both coordinate pairs stated clearly. Verified by substitution — a step Grade 7 students omit.

📋 Revision Sheet

📖 Key Definitions

Quadratic	Expression with highest power $x^2$
Discriminant	$b^2 - 4ac$ — determines number of roots
Completing the square	Writing $ax^2+bx+c$ as $a(x+p)^2+q$
Composite function	$fg(x) = f(g(x))$ — g applied first
Inverse function	$f^{-1}(x)$ — undoes $f$
Proof	A general algebraic argument, not examples

📐 Essential Formulae

Quadratic formula: $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

Perfect square: $(a+b)^2 = a^2 + 2ab + b^2$

Diff. of squares: $(a+b)(a-b) = a^2 - b^2$

Completing square: $(x+\tfrac{b}{2})^2 - (\tfrac{b}{2})^2 + c$

🧠 Memory Hooks

🦊 FOIL: First · Outer · Inner · Last (double brackets)
🤬 Negative → Flip: divide by negative = flip inequality
👈 fg = RIGHT first: in fg(x), g goes first
🔄 Inverse = Swap x and y
🔢 Even = 2n, Odd = 2n+1
⛰️ Turning point = (−p, q) from $(x+p)^2+q$ — OPPOSITE sign

🎯 Discriminant Quick Reference

$\Delta > 0$ → Two distinct real roots

$\Delta = 0$ → One repeated root (tangent)

$\Delta < 0$ → No real roots

Use in "show it has no real solutions" or "find k such that there are equal roots" questions.

🔄 Flashcards

Click any card to reveal the answer. Test yourself without looking at your notes.

✗ Common Mistakes

✗

Mistake 1 — Forgetting the middle term when squaring

Students write $(x+5)^2 = x^2 + 25$.
Correct: $(x+5)^2 = x^2 + 10x + 25$
Fix: Always expand using FOIL. Never "square each term separately".

✗

Mistake 2 — Wrong order for composite functions

Students evaluate $fg(x)$ by applying $f$ first, then $g$.
Correct: In $fg(x)$, apply $g$ first (the one nearest to $x$), then $f$.
Fix: Rewrite $fg(x) = f(\,g(x)\,)$ and substitute the inner function.

✗

Mistake 3 — Wrong turning point sign from completed square

Students read $y = (x+3)^2 + 2$ as having turning point $(3, 2)$.
Correct: Turning point is $(-3, 2)$ — the sign of $p$ reverses.
Fix: Set $x + 3 = 0$ to find the $x$-coordinate: $x = -3$.

✗

Mistake 4 — Algebraic proof using specific numbers

Students write: "Let $n = 3$. Then $2n+1 = 7$, which is odd. So it works."
Correct: This is a verification, not a proof. Use general algebra throughout.
Fix: Start with $n$ as a general integer. Never substitute a specific value in a proof.

✗

Mistake 5 — Quadratic inequality: single region instead of two

For $x^2 - 5x + 6 > 0$, students write $2 > x > 3$ (impossible region).
Correct: The parabola is above the x-axis outside the roots: $x < 2$ or $x > 3$.
Fix: Sketch the parabola. Shade where it satisfies the inequality. Read off regions.

✗

Mistake 6 — Simultaneous equations: finding x but forgetting y

Students stop after finding $x = 2$ and $x = -5$, forgetting to find both $y$ values.
Correct: Always find both $(x, y)$ pairs and state them as coordinate pairs.
Fix: Underline "find the coordinates" in the question. Coordinates = pairs.

✅ Final Checklist

Tick only when you can answer without looking at notes.

I understand what the discriminant means and can use it without the formula sheet
I understand why $fg(x) \neq gf(x)$ and can explain the difference
I understand why specific examples are not algebraic proofs
I can expand double brackets using FOIL without mistakes
I can factorise quadratics when $a \neq 1$ using the ac method
I can solve a quadratic using all three methods
I can complete the square for $ax^2 + bx + c$
I can calculate a composite function $fg(x)$ correctly
I can find an inverse function by swapping $x$ and $y$
I can solve a linear + quadratic pair of simultaneous equations and find both coordinate pairs
I can write a complete algebraic proof with a conclusion sentence
I can represent even, odd, and consecutive integers algebraically
I can solve a quadratic inequality and write the solution correctly
I can use the discriminant to find the value of $k$ for a given number of roots
I can find the minimum point of a parabola by completing the square

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