Edexcel International GCSE Maths Quadratic Equations Past Paper Questions and Solutions (4MA1 Higher)

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Edexcel International GCSE Maths A (4MA1) · Higher · Equations, Formulae and Identities

Quadratic Equations: Edexcel 4MA1 Past Paper Questions and Worked Solutions

This page collects every quadratic equations question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 8 questions worth 36 marks. Questions run from easier to harder. Try each one first, then click “Show solution” for a step-by-step method with the reason for every step. All answers are checked against the official Pearson mark schemes.

13 marksHigherJanuary 2020 · Paper 1HR (4MA1/1HR) · Q6

Solve x2 − 5x − 36 = 0. Show clear algebraic working.

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  1. Two numbers that multiply to −36 and add to −5: −9 and +4.
  2. (x − 9)(x + 4) = 0

Answer: x = 9 or x = −4

25 marksHigherJanuary 2021 · Paper 1HR (4MA1/1HR) · Q16

(a) Expand and simplify (x + 4)(x − 2)(x + 1) (3)

(b) Express x2 − 10x + 40 in the form (x + a)2 + b, where a and b are integers. (2)

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Part (a)

  1. (x + 4)(x − 2) = x2 + 2x − 8
  2. (x2 + 2x − 8)(x + 1) = x3 + x2 + 2x2 + 2x − 8x − 8

Answer: x3 + 3x2 − 6x − 8

Part (b)

  1. Half of −10 is −5: x2 − 10x = (x − 5)2 − 25
  2. (x − 5)2 − 25 + 40

Answer: (x − 5)2 + 15

36 marksHigherNovember 2021 · Paper 1H (4MA1/1H) · Q19

ABCED is a five-sided shape. ABCD is a rectangle with AB = x cm and BC = y cm. CED is an equilateral triangle on top of the rectangle. The perimeter of ABCED is 100 cm. The area of ABCED is R cm2.

ABCDEx cmy cm

(a) Show that R = x⁄4(200 − [6 − √3]x) (3)

(b)(i) Find the value of x for which R has its maximum value. Give your answer in the form p⁄q − √3 where p and q are integers. (2)

(ii) Explain why the maximum value of R is given by this value of x. (1)

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Part (a)

  1. The perimeter goes AB + BC + CE + ED + DA = x + y + x + x + y = 3x + 2y = 100, so y = 100 − 3x⁄2
  2. Area of the rectangle = xy = x(100 − 3x)⁄2 = x⁄4(200 − 6x)
  3. Area of an equilateral triangle with side x = ½x2 sin 60° = √3⁄4x2 = x⁄4(√3x)
  4. R = x⁄4(200 − 6x) + x⁄4(√3x) = x⁄4(200 − 6x + √3x)

Answer: R = x⁄4(200 − [6 − √3]x)

Part (b)(i)

  1. Expand: R = 50x − 6 − √3⁄4x2
  2. At the maximum, dR⁄dx = 0: 50 − 6 − √3⁄2x = 0
  3. x = 100⁄6 − √3

Answer: x = 100⁄6 − √3

Part (b)(ii)

  1. R is a quadratic in x with a negative x2 coefficient (6 − √3 is positive, so −6 − √3⁄4 is negative). Its graph is an upside-down U, so the turning point is a maximum.

Answer: The x2 coefficient is negative, so the stationary point is a maximum

44 marksHigherJanuary 2022 · Paper 1H (4MA1/1H) · Q20

(a) Express 7 + 12x − 3x2 in the form a + b(x + c)2 where a, b and c are integers. (3)

C is the curve with equation y = 7 + 12x − 3x2. The point A is the maximum point on C.

(b) Use your answer to part (a) to write down the coordinates of A. (1)

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Part (a)

  1. Take out −3 from the x terms: 7 − 3(x2 − 4x)
  2. Complete the square: x2 − 4x = (x − 2)2 − 4
  3. 7 − 3[(x − 2)2 − 4] = 7 + 12 − 3(x − 2)2

Answer: 19 − 3(x − 2)2, so a = 19, b = −3, c = −2

Part (b)

  1. −3(x − 2)2 is never positive, so y is largest (19) when x = 2.

Answer: (2, 19)

55 marksHigherJune 2024 · Paper 1HR (4MA1/1HR) · Q20

(a) Express 2x2 − 11x + 9 in the form a(x − b)2 − c where a, b and c are numbers to be found. (3)

The curve C has equation y = 2(x − 3)2 − 11(x − 3) + 9. The point P is the minimum point on C.

(b) Find the coordinates of P. (2)

Show solution

Part (a)

  1. Take out the factor 2 from the x terms: 2[x2 − 11⁄2x] + 9
  2. Complete the square: x2 − 11⁄2x = (x − 11⁄4)2 − (11⁄4)2 = (x − 11⁄4)2 − 121⁄16
  3. So 2[(x − 11⁄4)2 − 121⁄16] + 9 = 2(x − 11⁄4)2 − 121⁄8 + 9
  4. −121⁄8 + 9 = −121⁄8 + 72⁄8 = −49⁄8

Answer: 2(x − 2.75)2 − 6.125, i.e. a = 2, b = 11⁄4, c = 49⁄8

Part (b)

  1. From part (a), y = 2x2 − 11x + 9 has its minimum at (2.75, −6.125).
  2. The equation of C is the same expression with x replaced by (x − 3). That is a translation 3 units to the right.
  3. P = (2.75 + 3, −6.125)

Answer: (5.75, −6.125)

65 marksHigherJune 2022 · Paper 1HR (4MA1/1HR) · Q23

(a) Express 2x2 − 12x + 3 in the form a(x + b)2 + c where a, b and c are integers. (3)

The curve C has equation y = 2(x + 4)2 − 12(x + 4) + 3. The point M is the minimum point on C.

(b) Find the coordinates of M. (2)

Show solution

Part (a)

  1. Take out 2 from the x terms: 2(x2 − 6x) + 3
  2. Complete the square: x2 − 6x = (x − 3)2 − 9
  3. 2[(x − 3)2 − 9] + 3 = 2(x − 3)2 − 18 + 3

Answer: 2(x − 3)2 − 15

Part (b)

  1. y = 2x2 − 12x + 3 has its minimum at (3, −15).
  2. C is the same curve with x replaced by (x + 4), which is a translation 4 units to the left.

Answer: (−1, −15)

74 marksHigherJune 2022 · Paper 1H (4MA1/1H) · Q24

Express each of a, b and c in terms of q so that q + 12x − qx2 can be written as a − b(x − c)2

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  1. Take out −q from the x terms: q − q(x2 − 12⁄qx)
  2. Complete the square: x2 − 12⁄qx = (x − 6⁄q)2 − 36⁄q2
  3. q − q[(x − 6⁄q)2 − 36⁄q2] = q + 36⁄q − q(x − 6⁄q)2
  4. Compare with a − b(x − c)2.

Answer: a = q + 36⁄q, b = q, c = 6⁄q

84 marksHigherJune 2024 · Paper 1H (4MA1/1H) · Q25

f(x) = 17 − 3x2 + 12x. Write f(x) in the form a − b(x − c)2 where a, b and c are constants.

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  1. Group the x terms and take out −3: f(x) = 17 − 3(x2 − 4x)
  2. Complete the square inside the bracket: x2 − 4x = (x − 2)2 − 4
  3. f(x) = 17 − 3[(x − 2)2 − 4] = 17 − 3(x − 2)2 + 12
  4. Combine the numbers: 17 + 12 = 29

Answer: 29 − 3(x − 2)2

Tip: Check: expand 29 − 3(x2 − 4x + 4) = 29 − 3x2 + 12x − 12 = 17 − 3x2 + 12x ✓

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