What is A18: Quadratic equations?

Rearrange a quadratic equation into ax^2+bx+c=0 before choosing a method. Factorising is efficient when integer factors are visible and uses the zero-product rule.

How many marks are A18 questions worth?

In verified past-paper appearances, A18 questions on Edexcel GCSE Maths 1MA1 have been worth 3–7 marks.

Which past exam papers has A18 appeared in?

A18 has appeared in 26 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

A18

Solve quadratic equations (including those requiring rearrangement) algebraically by factorising, by completing the square and by using the quadratic formula; find approximate solutions using a graph

Quadratic equations

Worked answers, methods and verified real exam appearances for A18 on Edexcel GCSE Maths 1MA1.

Explanation

Rearrange a quadratic equation into $ax^2+bx+c=0$ before choosing a method. Factorising is efficient when integer factors are visible and uses the zero-product rule.
At Higher tier, completing the square rewrites the expression so a square can be isolated, while the quadratic formula $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ works generally.
Graphical roots are $x$ -intercepts, or intersection $x$ -coordinates when two graphs are compared.
The examiner expects both solutions unless the context rules one out.
Keep exact surd answers when asked and only round at the final step.

Worked example

Higher tier: Solve $x^2-8x+3=0$ by completing the square.

1.Rewrite the quadratic: $x^2-8x+3=(x-4)^2-13$ .
2.Set it equal to zero: $(x-4)^2=13$ .
3.Take both square roots: $x-4=\pm\sqrt{13}$ .

Answer: $x=4\pm\sqrt{13}$ .

Common mistakes

Don't take only the positive square root and lose one solution.
Don't substitute $b$ into the formula without its sign, so $-b$ is evaluated incorrectly.
Don't round a surd before the final answer and lose accuracy.

Exam tip

Write the substitution into the quadratic formula before evaluating; this is normally where the first method mark is earned.

Worked practice

Tier 1 · Easy

Solve

x^2+2x-35=0

by factorising.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$x=5$ or $x=-7$	2	$x^2+2x-35=(x+7)(x-5)$ . By the zero-product rule, $x+7=0$ or $x-5=0$ , so $x=-7$ or $x=5$ .

Tier 2 · Standard

Higher only: Solve

x^2+14x+5=0

by completing the square. Give exact answers.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$x=-7\pm2\sqrt{11}$	4	$x^2+14x+5=(x+7)^2-44$ . Hence $(x+7)^2=44$ , so $x+7=\pm\sqrt{44}=\pm2\sqrt{11}$ and $x=-7\pm2\sqrt{11}$ .

Tier 3 · Hard

Higher only: The curves

y=x^2

and

y=5x+1

intersect twice. Use the quadratic formula to find the exact

x

-coordinates, then give the values a graph should show to two decimal places.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$x=\frac{5\pm\sqrt{29}}{2}$ $x\approx-0.19$ or $x\approx5.19$	5	At an intersection, $x^2=5x+1$ , so $x^2-5x-1=0$ . The formula gives $x=\frac{5\pm\sqrt{(-5)^2-4(1)(-1)}}{2}=\frac{5\pm\sqrt{29}}{2}$ . These are approximately $-0.19$ and $5.19$ , matching the graph's intersection coordinates.

Tier 1 · Easy

Solve

x^2=64

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$x=-8$ or $x=8$	2	Both $8^2$ and $(-8)^2$ equal $64$ . Therefore $x=-8$ or $x=8$ .

Tier 2 · Standard

Solve

x^2-7x=44

by factorising.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$x=-4$ or $x=11$	3	Rearrange to get $x^2-7x-44=0$ . Factorise: $(x-11)(x+4)=0$ . Therefore $x=-4$ or $x=11$ .

Tier 3 · Hard

Solve

(x-4)(x+7)=12

by factorising.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$x=-8$ or $x=5$	3	Expand and rearrange: $x^2+3x-28=12$ , so $x^2+3x-40=0$ . Factorising gives $(x+8)(x-5)=0$ , hence $x=-8$ or $x=5$ .

Tier 2 · Standard

Solve

(x-3)^2=49

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$x=-4$ or $x=10$	3	Taking both square roots gives $x-3=7$ or $x-3=-7$ . Therefore $x=10$ or $x=-4$ .

Tier 3 · Hard

Higher only: Use the quadratic formula to solve

2x^2+5x-13=0

. Give your answers correct to

2

decimal places.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$x=1.59$ or $x=-4.09$	4	With $a=2$ , $b=5$ and $c=-13$ , the formula gives $x=\dfrac{-5\pm\sqrt{5^2-4(2)(-13)}}{4}=\dfrac{-5\pm\sqrt{129}}{4}$ . The values are $1.589454\ldots$ and $-4.089454\ldots$ , which round to $1.59$ and $-4.09$ . Each root is $0.004454\ldots$ , or $0.445$ final-digit units, from the nearest $2$ -decimal-place rounding boundary.

Tier 3 · Hard

Higher only: Solve

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

. Give both solutions correct to

2

decimal places.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$x=8.53$ or $x=0.47$	4	Expanding both sides gives $x^2+3x-4=2x^2-6x$ . Rearranging gives $x^2-9x+4=0$ . The quadratic formula gives $x=\dfrac{9\pm\sqrt{65}}{2}$ , so $x=8.5311\ldots$ or $x=0.4688\ldots$ , which round to $8.53$ and $0.47$ .

Q10

Tier 3 · Hard

Higher only: Solve

2x(x-3)=5-x

using the quadratic formula. Give both solutions in exact form.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$x=\dfrac{5\pm\sqrt{65}}{4}$	4	Expand and rearrange to obtain $2x^2-5x-5=0$ . With $a=2$ , $b=-5$ and $c=-5$ , the quadratic formula gives $x=\dfrac{5\pm\sqrt{(-5)^2-4(2)(-5)}}{4}=\dfrac{5\pm\sqrt{65}}{4}$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2019-11	1H	Q18	6	Non-calculator	Higher	QP MS
2022-11	1H	Q19	5	Non-calculator	Higher	QP MS
2021-11	2F	Q24	6	Allowed	Foundation	QP MS
2019-06	3H	Q20	5	Allowed	Higher	QP MS
2024-06	2H	Q5	6	Allowed	Higher	QP MS
2024-11	1F	Q26	3	Non-calculator	Foundation	QP MS
2023-06	1H	Q24	5	Non-calculator	Higher	QP MS
2019-11	3H	Q17	4	Allowed	Higher	QP MS
2022-06	1H	Q19	4	Non-calculator	Higher	QP MS
2024-11	2H	Q21	5	Allowed	Higher	QP MS
2019-11	1H	Q21	5	Non-calculator	Higher	QP MS
2021-11	1H	Q8	3	Non-calculator	Higher	QP MS
2022-06	3H	Q22	5	Allowed	Higher	QP MS
2019-11	1H	Q14	4	Non-calculator	Higher	QP MS
2021-11	2F	Q27	3	Allowed	Foundation	QP MS
2021-11	3H	Q16	4	Allowed	Higher	QP MS
2021-11	1H	Q22	4	Non-calculator	Higher	QP MS
2024-06	1H	Q21	5	Non-calculator	Higher	QP MS
2022-11	2H	Q19	3	Allowed	Higher	QP MS
2023-11	3H	Q17	7	Allowed	Higher	QP MS
2024-11	1H	Q13	5	Non-calculator	Higher	QP MS
2022-06	2H	Q22	5	Allowed	Higher	QP MS
2024-11	3H	Q20	4	Allowed	Higher	QP MS
2024-06	3H	Q20	7	Allowed	Higher	QP MS
2019-06	1H	Q17	4	Non-calculator	Higher	QP MS
2023-06	3H	Q21	4	Allowed	Higher	QP MS

Other points in A Algebra · sequences and functions

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