Pearson Edexcel International GCSE Mathematics A 4MA1: Quadratic equations – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 2.7, Quadratic equations, of the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on that specification. Questions 1 to 3 use Foundation tier content (solving x² + bx + c = 0 by factorisation) and suit both tiers. Questions marked (Higher) use content that appears only in the Higher tier: factorising ax² + bx + c, the quadratic formula, completing the square, forming equations from a context, and linear-quadratic simultaneous equations. A calculator may be used on every 4MA1 paper, so a calculator is allowed throughout, but show your algebra.

Before you start, you may want the quadratic equations study guide or the quadratic equations revision notes. For factorising and completing the square as expressions, use the algebraic manipulation practice questions. The course hub and the printable checklist cover the rest of 4MA1.

Questions

1. Solve x² + 9x + 20 = 0. [3]

2. Solve x² − 6x = 16. [3]

3. Solve y² = 11y. [2]

4. (Higher) Solve 3x² + 10x − 8 = 0 by factorising. [3]

5. (Higher) Solve 2x² − 5x − 4 = 0. Give your answers correct to 3 significant figures. [3]

6. (Higher)

(a) Write x² + 6x − 11 in the form (x + p)² + q, where p and q are integers. [2] (b) Hence solve x² + 6x − 11 = 0. Give your answers in the form a ± √b, where a and b are integers. [2]

7. (Higher) A right-angled triangle has shorter sides of length x cm and (x + 7) cm. Its hypotenuse has length (x + 8) cm.

(a) Show that x² − 2x − 15 = 0. [3] (b) Find the perimeter of the triangle. [2]

8. (Higher) Solve the simultaneous equations

y = 2x + 3 y = x² − 5

Show clear algebraic working. [5]

9. (Higher) The line y = x + 2 meets the circle x² + y² = 20 at the points A and B.

(a) Find the coordinates of A and B. Show clear algebraic working. [6] (b) Find the length of AB. Give your answer as a surd in its simplest form. [2]

10. (Higher) A rectangular lawn measures 12 m by 8 m. A path of constant width x m is laid all the way round the outside of the lawn. The total area of the lawn and the path is 150 m².

(a) Show that 2x² + 20x − 27 = 0. [3] (b) Find the width of the path. Give your answer correct to 3 significant figures. [3] (c) Paving for the path costs £18.50 per m². Work out the cost of paving the whole path. [2]

11. (Higher) Sana cycles 60 km at an average speed of x km/h. If she cycled 5 km/h faster, the journey would take 1 hour less.

(a) Show that x² + 5x − 300 = 0. [4] (b) Find the time Sana takes for the journey at her usual speed. [3]

Answers

1. Two numbers with product 20 and sum 9 are 4 and 5 [1]. So (x + 4)(x + 5) = 0 [1], giving x = −4 or x = −5 [1]. [3] Examiner insight: “Solve” needs values of x; correct brackets with no solutions stated usually earn only the method marks.

2. Rearrange to x² − 6x − 16 = 0 [1]. Factorise: (x − 8)(x + 2) = 0 [1]. So x = 8 or x = −2 [1]. [3] Examiner insight: The method mark comes from making one side zero first; factorising x² − 6x and setting each bracket equal to 16 earns nothing.

3. y² − 11y = 0, so y(y − 11) = 0 [1]. So y = 0 or y = 11 [1]. [2] Examiner insight: Dividing both sides by y gives only y = 11, so the answer is incomplete and loses the final mark.

4.

a x c = 3 x (-8) = -24; 12 and -2 multiply to -24 and add to 10
3x^2 + 12x - 2x - 8 = 3x(x + 4) - 2(x + 4)            [1]
(3x - 2)(x + 4) = 0                                     [1]

x = 2/3 or x = −4 [1] [3] Examiner insight: Brackets that give the right x² and constant terms but the wrong middle term earn no accuracy mark, so expand your brackets to check before solving.

5.

a = 2, b = -5, c = -4
x = ( 5 ± √((-5)^2 - 4(2)(-4)) ) / (2 x 2)              [1]
x = ( 5 ± √57 ) / 4                                     [1]

x = 3.14 or x = −0.637 [1] [3] Examiner insight: Writing the substituted formula earns the method mark even if a keying slip spoils the decimals; bare answers with no working risk scoring nothing if either value is wrong.

6. (a)

x^2 + 6x - 11 = (x + 3)^2 - 9 - 11                      [1]
             = (x + 3)^2 - 20                           [1]

p = 3, q = −20

(b)

(x + 3)^2 = 20
x + 3 = ±√20                                            [1]

x = −3 ± √20 [1] (−3 ± 2√5 is equivalent and also correct). [4] Examiner insight: “Hence” means you must use part (a); solving (b) with the formula instead may not gain the marks, and dropping the ± loses the final accuracy mark.

7. (a)

By Pythagoras: x^2 + (x + 7)^2 = (x + 8)^2              [1]
x^2 + x^2 + 14x + 49 = x^2 + 16x + 64                   [1]
2x^2 + 14x + 49 - x^2 - 16x - 64 = 0
x^2 - 2x - 15 = 0                                       [1]

(b) (x − 5)(x + 3) = 0, so x = 5 or x = −3; a length cannot be negative, so x = 5 [1]. The sides are 5 cm, 12 cm and 13 cm, so the perimeter is 30 cm [1]. [5] Examiner insight: In a “show that” the answer is given, so every line of expansion must be seen; jumping from the Pythagoras statement to the given equation scores only the first mark.

8.

x^2 - 5 = 2x + 3                                        [1]
x^2 - 2x - 8 = 0                                        [1]
(x - 4)(x + 2) = 0                                      [1]
x = 4  or  x = -2                                       [1]
y = 2(4) + 3 = 11;  y = 2(-2) + 3 = -1                  [1]

x = 4, y = 11 and x = −2, y = −1 [5] Examiner insight: With “show clear algebraic working”, correct pairs from a graph or a calculator with no algebra can score zero, and the last mark needs the values correctly paired.

9. (a)

x^2 + (x + 2)^2 = 20                                    [1]
x^2 + x^2 + 4x + 4 = 20                                 [1]
2x^2 + 4x - 16 = 0,  so  x^2 + 2x - 8 = 0               [1]
(x + 4)(x - 2) = 0                                      [1]
x = 2  or  x = -4                                       [1]
x = 2 -> y = 4;  x = -4 -> y = -2                       [1]

A(2, 4) and B(−4, −2) (either order)

(b)

AB^2 = (2 - (-4))^2 + (4 - (-2))^2 = 36 + 36 = 72       [1]

AB = √72 = 6√2 [1] (8.49 to 3 s.f.) [8] Examiner insight: Part (b) is usually marked with follow-through from your points in (a), so set out the Pythagoras step clearly even if you doubt your coordinates.

10. (a)

Outer rectangle: (12 + 2x) by (8 + 2x)
(12 + 2x)(8 + 2x) = 150                                 [1]
96 + 24x + 16x + 4x^2 = 150                             [1]
4x^2 + 40x - 54 = 0,  divide by 2:  2x^2 + 20x - 27 = 0 [1]

(b)

x = ( -20 ± √(20^2 - 4(2)(-27)) ) / (2 x 2)             [1]
x = ( -20 ± √616 ) / 4                                  [1]
x = 1.2048...  or  x = -11.20...

A width cannot be negative, so the path is 1.20 m wide [1].

(c) Path area = 150 − 12 × 8 = 54 m² [1]. Cost = 54 × 18.50 = £999 [1]. [8] Examiner insight: The path adds x to each side, so the outer lengths are 12 + 2x and 8 + 2x; using 12 + x and 8 + x loses every mark in (a).

11. (a)

Usual time = 60/x hours                                 [1]
Faster time = 60/(x + 5) hours                          [1]
60/x - 60/(x + 5) = 1                                   [1]
Multiply by x(x + 5): 60(x + 5) - 60x = x(x + 5)
300 = x^2 + 5x,  so  x^2 + 5x - 300 = 0                 [1]

(b) (x + 20)(x − 15) = 0 [1]. A speed cannot be negative, so x = 15 [1]. Time = 60 ÷ 15 = 4 hours [1]. [7] Examiner insight: The question asks for a time, so giving x = 15 as the final answer loses the last mark; also multiply the right-hand 1 by x(x + 5), not just the fractions.

Where marks are usually lost

• Not rearranging to = 0 before factorising, as in setting (x − 8) equal to 16.
• Writing the correct brackets but not stating the solutions when the question says “solve”.
• Dividing through by the variable and losing the solution 0, as in Question 3.
• Taking x = 2 instead of x = 2/3 from the bracket (3x − 2).
• Mis-substituting a negative b or c into the quadratic formula, especially −(−5) and −4(2)(−4).
• Ignoring “hence” in Question 6 and losing the method marks.
• In “show that” questions, skipping lines of expansion or starting from the given answer.
• Expanding (x + 2)² as x² + 4 when substituting into a circle equation.
• Keeping a negative length, width or speed in a context answer instead of rejecting it with a reason.
• Giving x when the question asked for a perimeter, a cost or a time.

Next steps

• Go back over any weak method in the quadratic equations revision notes.
• Re-read the full explanations in the quadratic equations study guide.
• Revise factorising expressions with the use of symbols and algebraic manipulation study guide.
• See every topic on the Edexcel IGCSE Mathematics course hub.
• Tick off what you know on the printable checklist.
• Try one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification Issue 2, November 2017, first examination June 2018, published by Pearson Education Limited. Section 2.7 Quadratic equations (Foundation tier statement A; Higher tier statements A to D).