A18, A11AlgebraFoundation & HigherGrades 4–8

Quadratic Equations

Q: How do I choose a method?

Try factorising first if the numbers are small and the question does not mention decimals or surds. If it asks for decimal places, use the formula. If it asks for surd form or the turning point, completing the square is often neatest.

Q: What does b^2 - 4ac tell me?

It is called the discriminant. Positive: two solutions. Zero: one repeated solution. Negative: no real solutions (the graph never meets the x-axis).

Q: Can a quadratic have only one solution?

Yes, when it is a perfect square such as (x - 4)^2 = 0. The graph just touches the x-axis at x = 4.

A quadratic equation has an $x^2$ term and no higher power, such as $x^2 - 5x + 6 = 0$ . Unlike a linear equation it usually has two solutions, because a U-shaped graph can cross the $x$ -axis twice. There are three methods, and choosing the right one quickly is a big part of the skill.

55 min Video by Cognito AQA GCSE Maths

Before you start

Expand two brackets, substitute negative numbers and recognise an equation in the form ax² + bx + c = 0.

Try this first. If you need the answer, revisit the prerequisite before continuing.

Starting check: Expand (x − 4)(x + 2).

Show answer and reasoning

x² − 2x − 8. The middle terms are 2x and −4x.

The big picture

Every method relies on getting the equation into the form $ax^2 + bx + c = 0$ first. The key fact is that if two things multiply to give zero, at least one of them must be zero — so a factorised quadratic can be split into two simple linear equations.

Factorising is fastest when it works. Completing the square also finds the turning point of the graph. The quadratic formula works for every quadratic, and is the method to use when the question asks for answers to decimal places — a strong hint that it will not factorise.

GCSE Maths - Solving Quadratics Using the Quadratic Formula (2026/27 exams)Curated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

Watch, pause, try

Video by Cognito. This clip teaches the quadratic formula, generally Higher-tier content. Use the tier guidance and factorising examples in your course before choosing a method.

Viewing focus: Identify a, b and c with their signs, and keep the entire numerator over 2a.

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: For 2x² − 3x − 4 = 0, identify a, b and c, then calculate b² − 4ac.

Show answer and reasoning

a = 2, b = −3 and c = −4. The discriminant is 9 + 32 = 41.

After watching 2: Use the formula to give the exact roots of $2x^2-3x-4=0$ .

Show answer and reasoning

$x=\frac{3\pm\sqrt{41}}4$ . The two approximate roots are 2.35 and −0.851.

KEPLER VISUAL LAB

Explore the idea

Predict what will change, then use the controls to test your explanation.

Point on the curve: x1

At x = 1, y = -12. A root makes y exactly zero: either x + 3 = 0 or x − 4 = 0.

On a narrow screen, scroll the diagram sideways to read every label.

What you'll learn

Rearrange a quadratic equation into the form $ax^2 + bx + c = 0$
Solve quadratic equations by factorising
Solve by completing the square (Higher)
Solve using the quadratic formula (Higher)

Solving by factorising

Rearrange so one side is zero. Factorise, then set each bracket equal to zero and solve.

Two special cases: if there is no constant term, take out $x$ as a common factor — never divide by $x$ , or you lose the solution $x = 0$ . If there is no $x$ term, as in $x^2 = 49$ , square root and remember $\pm$ .

Example 1Solve

x^2 + 2x = 15

1Rearrange:

x^2 + 2x - 15 = 0

2Factorise:

(x + 5)(x - 3) = 0

3So

x + 5 = 0

x - 3 = 0

Answer

x = -5

x = 3

Your turnSame method, new numbers — try it before moving on.

Solve $x^2 - 7x + 10 = 0$ .

Your answer

Separate the solutions with a comma.

Show worked solution

Solution

1Multiply to 10, add to $-7$ : $-2$ and $-5$ .
2 $(x - 2)(x - 5) = 0$ , so $x = 2$ or $x = 5$ .

Answer: $x = 2$ or $x = 5$

Example 2Solve

3x^2 = 12x

1Rearrange:

3x^2 - 12x = 0

2Factorise:

3x(x - 4) = 0

3x = 0

x - 4 = 0

Answer

x = 0

x = 4

Your turnSame method, new numbers — try it before moving on.

Solve $2x^2 + 5x - 3 = 0$ .

Your answer

Separate with a comma; 1/2 or 0.5 are both fine.

Show hint

Factorise as $(2x - 1)(x + 3)$ — check by expanding.

Show worked solution

Solution

1 $(2x - 1)(x + 3) = 0$ .
2 $2x - 1 = 0$ gives $x = \frac{1}{2}$ ; $x + 3 = 0$ gives $x = -3$ .

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

Tip — The solutions have the opposite sign to the numbers in the brackets: $(x + 5)$ gives $x = -5$ .

Completing the square (Higher)

Rewrite $x^2 + bx + c$ as $\left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c$ . Then the equation can be solved by rearranging and square rooting.

If the $x^2$ coefficient is not 1, take it out as a factor from the $x^2$ and $x$ terms first.

Example 1Solve

x^2 + 6x - 4 = 0

, leaving your answers in surd form.

1Complete the square:

(x + 3)^2 - 9 - 4 = 0

, so

(x + 3)^2 = 13

2Square root:

x + 3 = \pm\sqrt{13}

3Subtract 3.

Answer

x = -3 \pm \sqrt{13}

Your turnSame method, new numbers — try it before moving on.

Write $x^2 - 10x + 7$ in the form $(x - a)^2 - b$ .

Your answer

Type it like (x-3)^2-4.

Show worked solution

Solution

1Half of $-10$ is $-5$ : $(x - 5)^2 = x^2 - 10x + 25$ .
2 $x^2 - 10x + 7 = (x - 5)^2 - 25 + 7 = (x - 5)^2 - 18$ .

Answer: $(x - 5)^2 - 18$

Why the turning point appears

$(x + 3)^2 - 13$ can never be less than $-13$ , because a square is never negative. So the graph of $y = x^2 + 6x - 4$ has its lowest point at $(-3, -13)$ . Completing the square solves the equation and finds the turning point in one go.

The quadratic formula (Higher)

The solutions of $ax^2 + bx + c = 0$ are given by the quadratic formula, shown below.

Write down $a$ , $b$ and $c$ with their signs before substituting, and put negative values in brackets. Work out $b^2 - 4ac$ first — if it is negative, there are no real solutions.

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

The one method that works for every quadratic — know it by heart.

Example 1Solve

2x^2 - 3x - 7 = 0

, giving your answers to 2 decimal places.

a = 2

b = -3

c = -7

b^2 - 4ac = (-3)^2 - 4(2)(-7) = 9 + 56 = 65

x = \dfrac{3 \pm \sqrt{65}}{4}

, giving

2.766\ldots

and

-1.265\ldots

Answer

x = 2.77

x = -1.27

Your turnSame method, new numbers — try it before moving on.

Solve $x^2 + 4x - 3 = 0$ , giving your answers to 2 decimal places.

Your answer

Separate with a comma, to 2 d.p.

Show hint

$b^2 - 4ac = 16 + 12 = 28$ .

Show worked solution

Solution

1 $a = 1$ , $b = 4$ , $c = -3$ : $b^2 - 4ac = 28$ .
2 $x = \dfrac{-4 \pm \sqrt{28}}{2}$ .
3 $x = 0.6457\ldots$ or $x = -4.6457\ldots$

Answer: $x = 0.65$ or $x = -4.65$

Tip — The formula has $2a$ under the whole numerator — draw a long fraction line so the $-b$ is divided too.

Forming quadratic equations

Area problems and number problems often lead to quadratics. Form the equation, rearrange to $= 0$ , solve, and then reject any solution that does not make sense — a length cannot be negative.

Example 1A rectangle is 3 cm longer than it is wide. Its area is 40 cm². Find its width.

1Width

x

, length

x + 3

x(x + 3) = 40

x^2 + 3x - 40 = 0

, so

(x + 8)(x - 5) = 0

x = -8

is impossible for a length.

Answer

The width is 5 cm.

Your turnSame method, new numbers — try it before moving on.

The product of two consecutive positive integers is 132. What is the smaller integer?

Your answer

Show hint

Let them be $n$ and $n + 1$ , so $n(n + 1) = 132$ .

Show worked solution

Solution

1 $n^2 + n - 132 = 0$ .
2 $(n + 12)(n - 11) = 0$ , so $n = 11$ (reject $-12$ ).

Answer: $11$

Think like an examiner

"Give your answers to 2 decimal places" means use the formula (or completing the square) — do not waste time trying to factorise.
Show the substitution into the formula in full. A correct substitution earns a method mark even if the calculator work slips.
If you get $b^2 - 4ac < 0$ on an exam question, recheck your signs — GCSE questions almost always have real solutions.

Methods

pq = 0 \Rightarrow p = 0 \text{ or } q = 0

Why factorising works.

x^2 + bx = \left(x + \tfrac{b}{2}\right)^2 - \left(\tfrac{b}{2}\right)^2

Completing the square.

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

The quadratic formula — learn it.

Watch out for these

Solving

x^2 = 5x

by dividing by

x

to get

x = 5

Factorise

x(x - 5) = 0

x = 0

x = 5

Solving

x^2 = 36

x = 6

only.

x = \pm 6

(-3)^2

typed as

-3^2 = -9

in the formula.

Use brackets:

(-3)^2 = 9

Factorising without rearranging:

x^2 + x = 6 \Rightarrow x(x + 1) = 6

, so

x = 6

x = -1

Make one side zero first:

(x + 3)(x - 2) = 0

Stretch yourself

A right-angled triangle has sides $x$ cm, $(x + 7)$ cm and hypotenuse $(x + 8)$ cm. Find the perimeter of the triangle.

Hint — Use Pythagoras to form a quadratic equation, then solve it.

Reveal solution

Solution

x^2 + (x + 7)^2 = (x + 8)^2

gives

x^2 + x^2 + 14x + 49 = x^2 + 16x + 64

, so

x^2 - 2x - 15 = 0

. Then

(x - 5)(x + 3) = 0

, and

x = 5

(a length cannot be

-3

). The sides are 5, 12 and 13 cm, so the perimeter is 30 cm.

\blacksquare

Questions students ask

How do I choose a method?

Try factorising first if the numbers are small and the question does not mention decimals or surds. If it asks for decimal places, use the formula. If it asks for surd form or the turning point, completing the square is often neatest.

What does

b^2 - 4ac

tell me?

It is called the discriminant. Positive: two solutions. Zero: one repeated solution. Negative: no real solutions (the graph never meets the

x

-axis).

Can a quadratic have only one solution?

Yes, when it is a perfect square such as

(x - 4)^2 = 0

. The graph just touches the

x

-axis at

x = 4

Key takeaways

Rearrange to $ax^2 + bx + c = 0$ before doing anything else.
Factorise and set each factor to zero.
Never divide by $x$ ; take it out as a factor.
Completing the square gives exact (surd) answers and the turning point.
The quadratic formula works every time — learn it and substitute carefully.

Exam-style questions

Practise it the way it is examined

3 original questions · 11 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 43 marks

(a)Factorise $x^2 - 3x - 28$ .[2]
Show mark scheme (a)
- M1 $(x \pm 7)(x \pm 4)$
- A1 $(x - 7)(x + 4)$
Answer: $(x - 7)(x + 4)$
(b)Hence, or otherwise, solve $x^2 - 3x - 28 = 0$ .[1]
Show mark scheme (b)
- B1 $x = 7$ and $x = -4$ (follow through from their factors)
Answer: $x = 7$ or $x = -4$

Question 2Grade 63 marks

Solve $3x^2 + 5x - 1 = 0$ . Give your answers to 2 decimal places.[3]
Show mark scheme
- M1Correct substitution into the formula: $\dfrac{-5 \pm \sqrt{5^2 - 4 \times 3 \times (-1)}}{2 \times 3}$
- M1Simplifies the discriminant: $\sqrt{37}$
- A1 $x = 0.18$ and $x = -1.85$
Answer: $x = 0.18$ or $x = -1.85$

Question 3Grade 85 marks

The diagram shows a path of width $x$ metres around a rectangular lawn 8 m by 5 m. The total area of the lawn and path is 70 m².

The path has the same width

x

on every side.

(a)Show that $4x^2 + 26x - 30 = 0$ .[3]
Show mark scheme (a)
- M1Total dimensions $(8 + 2x)$ and $(5 + 2x)$
- M1 $(8 + 2x)(5 + 2x) = 70$ expanded: $40 + 16x + 10x + 4x^2 = 70$
- A1Rearranges to $4x^2 + 26x - 30 = 0$ with no errors
(b)Find the width of the path.[2]
Show mark scheme (b)
- M1Divides by 2 to get $2x^2 + 13x - 15 = 0$ and factorises, $(2x + 15)(x - 1) = 0$ , or uses the formula
- A1 $x = 1$ metre, with $x = -7.5$ rejected
Answer: $1$ m

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Complete the method

Some of the setup is done. Finish it on paper, then compare your reasoning.

Your turn: Complete: x² − x − 12 = 0 becomes (x − ___)(x + ___) = 0. Hence find both roots.

Show a hint

Find two numbers whose product is −12 and sum is −1.

Show answer and reasoning

The gaps are 4 and 3. Either x − 4 = 0 or x + 3 = 0, giving x = 4 or x = −3.

Spot and repair the mistake

Identify the first incorrect step and explain how to fix it.

Diagnose it: A student divides x² = 5x by x and reports only x = 5. What solution did they lose?

Show answer and reasoning

x = 0 is also a solution. Dividing by x excludes it. Instead write x² − 5x = x(x − 5) = 0, so x = 0 or x = 5.

Apply it in an exam

Write a full response before revealing the solution. These are Kepler-authored exam-style questions; the suggested marks are a self-check guide.

Question 1 · 4 marks: A rectangle has width x cm and length (x + 3) cm. Its area is 40 cm². Form an equation and find its dimensions.

Show answer and reasoning

x(x + 3) = 40, so x² + 3x − 40 = (x + 8)(x − 5) = 0. The roots are −8 and 5. A length must be positive, giving width 5 cm and length 8 cm.

Form x(x + 3) = 40.
Rearrange and factorise.
Find both algebraic roots.
Reject the negative length and state both dimensions.

Explore further

AQA 8300 specification (opens in a new tab)
Check the assessed content, tier requirements and official course documents.
Corbettmaths: quadratic formula practice (opens in a new tab)
Additional formula questions; check your tier before attempting these.

Recall and check later

Return with the notes closed. Answer from memory, check the reasoning, then practise anything you missed.

Recall 1: Tomorrow: solve x² + 2x − 15 = 0.

Show answer and reasoning

(x + 5)(x − 3) = 0, so x = −5 or 3.

Recall 2: In a week: why must an equation equal zero before using the zero-product rule?

Show answer and reasoning

A product equals zero only if at least one factor equals zero. A nonzero right-hand side does not let you set individual factors to zero.

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

Solve $(x - 4)(x + 9) = 0$ .
- $x = 4$ or $x = -9$
- $x = -4$ or $x = 9$
- $x = 4$ or $x = 9$
- $x = 36$
Show answer and explanation
Answer: $x = 4$ or $x = -9$
Set each bracket to zero.
Solve $x^2 = 81$ .
- $x = \pm 9$
- $x = 9$
- $x = 40.5$
- $x = \pm 40.5$
Show answer and explanation
Answer: $x = \pm 9$
Both $9^2$ and $(-9)^2$ equal 81.
Solve $x^2 + 5x + 6 = 0$ .
- $x = -2$ or $x = -3$
- $x = 2$ or $x = 3$
- $x = -1$ or $x = -6$
- $x = 1$ or $x = 6$
Show answer and explanation
Answer: $x = -2$ or $x = -3$
$(x + 2)(x + 3) = 0$ .
Solve $x^2 - 6x = 0$ .
- $x = 0$ or $x = 6$
- $x = 6$
- $x = \pm 6$
- $x = 0$ or $x = -6$
Show answer and explanation
Answer: $x = 0$ or $x = 6$
$x(x - 6) = 0$ .
Solve $x^2 - x = 12$ .
- $x = 4$ or $x = -3$
- $x = -4$ or $x = 3$
- $x = 12$ or $x = 13$
- $x = 6$ or $x = -2$
Show answer and explanation
Answer: $x = 4$ or $x = -3$
$x^2 - x - 12 = (x - 4)(x + 3) = 0$ .
For $2x^2 - 5x + 1 = 0$ , what is $b^2 - 4ac$ ?
- $17$
- $33$
- $-17$
- $23$
Show answer and explanation
Answer: $17$
$25 - 8 = 17$ .
Solve $x^2 + 4x - 1 = 0$ in surd form.
- $x = -2 \pm \sqrt{5}$
- $x = 2 \pm \sqrt{5}$
- $x = -2 \pm \sqrt{3}$
- $x = -4 \pm \sqrt{5}$
Show answer and explanation
Answer: $x = -2 \pm \sqrt{5}$
$(x + 2)^2 - 5 = 0$ .
How many real solutions does $x^2 + 2x + 5 = 0$ have?
- $0$
- $1$
- $2$
- $3$
Show answer and explanation
Answer: $0$
$b^2 - 4ac = 4 - 20 = -16 < 0$ .

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