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IGCSE · 4MA1 · Topic

Completing the Square: past paper questions & worked solutions

Every Completing the Square question from the 4MA1 papers in our bank, 2020–2026, sorted from easiest to hardest. Try each one first, then open the worked solution.

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18questions
70marks in total
31%of papers (17 of 54) include it
1/12/5easy / medium / hard

Questions, easiest first

Level is estimated from marks: 1–2 easy, 3–4 medium, 5+ hard.

1. Nov 2025 4WM1H · Q22

Easy2 marks
Nov 2025 4WM1H Question 22 (2 marks): Completing the Square – 22 9 12 2 x x q can be written in the form px r 2 where p, q and r are integers. Find the value of q q = (Total for Question 22 is 2 marks)
Show worked solution
  1. Calculate value
    \[ 9x^2-12x+q=(3x-2)^2 \]
    \[ (3x-2)^2=9x^2-12x+4 \]
Final answer

\(q=4\).

2. May 2026 4WM1H · Q24

Medium3 marks
May 2026 4WM1H Question 24 (3 marks): Completing the Square – 24 Express 22 3 30 2 x x in the form a b x c − − ( )2 where a, b and c are integers. (Total for Question 24 is 3 marks)
Show worked solution
  1. Factor out −3 from the x terms
    \[22-3x^2+30x=22-3(x^2-10x)\]
  2. Complete the square
    \[x^2-10x=(x-5)^2-25\]
  3. Simplify
    \[22-3\big((x-5)^2-25\big)=97-3(x-5)^2\]
Final answer

\(97-3(x-5)^2\)

3. Nov 2025 P1H · Q21

Medium3 marks
Nov 2025 P1H Question 21 (3 marks): Completing the Square – 21 Express 5x 2 ‒ 20x + 23 in the form a(x - b) 2 + c where a, b and c are integers. (Total for Question 21 is 3 marks)
Show worked solution
  1. Solve inequality
    \[ 5x^2-20x+23=5(x^2-4x)+23 \]
    \[ =5\left((x-2)^2-4\right)+23 \]
    \[ =5(x-2)^2-20+23 \]
    \[ =5(x-2)^2+3 \]
Final answer

\(5(x-2)^2+3\)

4. Nov 2025 4WM1H · Q24

Medium3 marks
Nov 2025 4WM1H Question 24 (3 marks): Completing the Square – 24 Express 5x2 ‒ 20x + 23 in the form a(x – b)2 + c where a, b and c are integers. (Total for Question 24 is 3 marks)
Show worked solution
  1. Calculate value
    \[ 5x^2-20x+23=5(x^2-4x)+23 \]
    \[ =5\big((x-2)^2-4\big)+23 \]
    \[ =5(x-2)^2+3 \]
Final answer

\(5(x-2)^2+3\).

5. Jun 2023 P2H · Q19

Medium3 marks
Jun 2023 P2H Question 19 (3 marks): Completing the Square – 19 Express 3x2 - 6x + 5 in the form a(x - b)2 + c (Total for Question 19 is 3 marks)
Show worked solution
  1. Calculate value
    \[ 3x^2-6x+5=3(x^2-2x)+5 \]
    \[ =3\big((x-1)^2-1\big)+5 \]
    \[ =3(x-1)^2+2 \]
Final answer

\(3(x-1)^2+2\)

6. Jan 2022 P1HR · Q19

Medium3 marks
Jan 2022 P1HR Question 19 (3 marks): Completing the Square – 19 Given that 1 3 x xm    (a) find the value of m m = (1) Given that a, b and c are integers, (b) express 3x2 + 12x + 19 in the form a(x + b)2 + c (2) (Total for Question 19 is 3 marks)
Show worked solution
  1. (a)
    \[ \sqrt[3]{\frac1x}=x^{-1/3} \]
  2. Evaluate fraction

    So \(m=-\frac13\).

  3. (b)
    \[ 3x^2+12x+19=3(x^2+4x)+19 \]
    \[ =3\big((x+2)^2-4\big)+19 \]
    \[ =3(x+2)^2+7 \]
Final answer

\(m=-\frac13\), and \(3(x+2)^2+7\)

7. Jan 2020 P2HR · Q23

Medium3 marks
Jan 2020 P2HR Question 23 (3 marks): Completing the Square – 23 Express 7 - 12x - 2x2 in the form a + b(x + c)2 where a, b and c are integers. (Total for Question 23 is 3 marks)
Show worked solution
  1. Solve inequality
    \[ 7-12x-2x^2=-2x^2-12x+7 \]
    \[ =-2(x^2+6x)+7 \]
    \[ =-2\left((x+3)^2-9\right)+7 \]
    \[ =-2(x+3)^2+18+7 \]
    \[ =25-2(x+3)^2 \]
Final answer

\(25-2(x+3)^2\)

8. May 2025 4WM1HR · Q18

Medium4 marks
May 2025 4WM1HR Question 18 (4 marks): Completing the Square – 18 (a) Express 2 12 23 2 x x in the form a x b c 2 where a, b and c are integers. Show clear algebraic working. (3) (b) Hence, state the least value of 2 12 23 2 x x (1) (Total for Question 18 is 4 marks)
Show worked solution
  1. Calculate value
    \[ 2x^2-12x+23=2(x^2-6x)+23 \]
    \[ =2\big((x-3)^2-9\big)+23 \]
    \[ =2(x-3)^2+5 \]
  2. The squared part is at

    The squared part is at least zero.

Final answer

\(2(x-3)^2+5\); least value \(5\).

9. May 2024 P1H · Q25

Medium4 marks
May 2024 P1H Question 25 (4 marks): Completing the Square – 25 f (x) = 17 - 3x 2 + 12x Write f (x) in the form a - b(x - c) 2 where a, b and c are constants. f (x) = (Total for Question 25 is 4 marks)
Show worked solution
  1. Solve inequality
    \[ f(x)=17-3x^2+12x \]
    \[ =-3x^2+12x+17 \]
    \[ =-3(x^2-4x)+17 \]
    \[ =-3\left((x-2)^2-4\right)+17 \]
    \[ =-3(x-2)^2+12+17 \]
    \[ =29-3(x-2)^2 \]
Final answer

\(f(x)=29-3(x-2)^2\)

10. Jun 2022 P1H · Q24

Medium4 marks
Jun 2022 P1H Question 24 (4 marks): Completing the Square – 24 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 a = b = c = (Total for Question 24 is 4 marks)
Show worked solution
  1. Evaluate fraction
    \[ q+12x-qx^2=-q\left(x^2-\frac{12}{q}x\right)+q \]
  2. Complete the square

    Complete the square:

    \[ x^2-\frac{12}{q}x=\left(x-\frac6q\right)^2-\frac{36}{q^2} \]
  3. Evaluate fraction

    So

    \[ q+12x-qx^2 =-q\left[\left(x-\frac6q\right)^2-\frac{36}{q^2}\right]+q \]
    \[ =q+\frac{36}{q}-q\left(x-\frac6q\right)^2 \]
  4. Compare coefficients

    Compare with

    \[ a-b(x-c)^2 \]
Final answer

\(a=q+\dfrac{36}{q}\), \(b=q\), \(c=\dfrac6q\)

11. Jan 2022 P1H · Q20

Medium4 marks
Jan 2022 P1H Question 20 (4 marks): Completing the Square – 20 (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 do
Show worked solution
  1. (a)
    \[ 7+12x-3x^2=-3x^2+12x+7 \]
    \[ =-3(x^2-4x)+7 \]
    \[ =-3\left((x-2)^2-4\right)+7 \]
    \[ =19-3(x-2)^2 \]
  2. This is the same as
    \[ 19-3(x+(-2))^2 \]
  3. (b) The maximum point is

    (b) The maximum point is read from the completed square form.

    \[ y=19-3(x-2)^2 \]
  4. Calculate value

    The maximum occurs when \(x=2\), giving \(y=19\).

Final answer

(a) \(19-3(x-2)^2\), (b) \(A=(2,19)\)

12. May 2021 P2H · Q22

Medium4 marks
May 2021 P2H Question 22 (4 marks): Completing the Square – 22 The curve S has equation y = f(x) where f(x) = x2 The curve T has equation y = g(x) where g(x) = 2x2 - 12x + 13 By writing g(x) in the form a(x - b)2 - c, where a, b and c are constants, describe fully a series of tra
Show worked solution
  1. Solve inequality
    \[ g(x)=2x^2-12x+13 \]
    \[ =2(x^2-6x)+13 \]
    \[ =2\left((x-3)^2-9\right)+13 \]
    \[ =2(x-3)^2-18+13 \]
    \[ =2(x-3)^2-5 \]
  2. Calculate value

    Starting with \(S: y=x^2\):

  3. Translate 3 units to the

    - Translate 3 units to the right to get \(y=(x-3)^2\). - Stretch parallel to the \(y\)-axis by scale factor 2 to get \(y=2(x-3)^2\). - Translate 5 units down to get \(y=2(x-3)^2-5\).

Final answer

\(g(x)=2(x-3)^2-5\). Transformations: right 3, stretch parallel to the \(y\)-axis by scale factor 2, then down 5.

13. Specimen 4WM1H · Q20

Medium4 marks
Specimen 4WM1H Question 20 (4 marks): Completing the Square – 20 Find a, b and c so that 7 + 12x - 2x^2 is written as a - b(x - c)^2. (Total for Question 20 is 4 marks)
Show worked solution
  1. Factor out -2 from the quadratic terms
    \[ 7+12x-2x^2=-2(x^2-6x)+7 \]
  2. Complete the square
    \[ x^2-6x=(x-3)^2-9 \]
  3. Simplify the expression
    \[ -2((x-3)^2-9)+7 =-2(x-3)^2+18+7 \]
    \[ =25-2(x-3)^2 \]
  4. Read the values
    \[ a=25,\qquad b=2,\qquad c=3 \]
Final answer

\(a=25,\ b=2,\ c=3\).

14. Jun 2026 P1H · Q24

Hard5 marks
Jun 2026 P1H Question 24 (5 marks): Completing the Square – 24 (a) Express 10 24 3 2 − − x x in the form a b x c � � � � 2 where a, b and c are integers. (3) T is the curve with equation y x x � � � � �� � � � 10 24 1 3 1 2 The point P is the maximum point on T (b) Use your ans
Show worked solution
  1. Part (a) | Step 1
    \[10-24x-3x^2=10-3(x^2+8x).\]

    Factor -3 from the terms involving x.

  2. Part (a) | Step 2
    \[x^2+8x=(x+4)^2-16.\]

    Half the coefficient of x is 4.

  3. Part (a) | Step 3
    \[10-3\big((x+4)^2-16\big)=58-3(x+4)^2.\]

    Thus a=58, b=3 and c=4.

  4. Part (b) | Step 1
    \[y=58-3\big((x-1)+4\big)^2=58-3(x+3)^2.\]

    Use (a) with x replaced throughout by x-1.

  5. Part (b) | Step 2

    Since \((x+3)^2\ge0\), the greatest y is 58, attained at \(x=-3\). Thus \(P=(-3,58)\).

    The negative square term is largest when its square is zero.

  6. Part (c) | Step 1

    The parabola is symmetric about its maximum, so its line of symmetry is \(x=-3\).

    In 58-3(x+3)^2, equal positive and negative offsets from -3 give the same y.

Final answer

(a) \(58-3(x+4)^2\)

(b) \(P=(-3,58)\)

(c) \(x=-3\)

15. May 2025 4WM1H · Q25

Hard5 marks
May 2025 4WM1H Question 25 (5 marks): Completing the Square – 25 (a) Express 8 12 4 2 − − x x in the form a b x c ( )2 where a, b and c are constants. (3) (b) Hence solve 8 12 4 0 2 x x Show your working clearly. Give your solutions in surd form. (2) (Total for Question 25 is 5
Show worked solution
  1. Simplify surd
    \[ 8x^2-12x-4=8\left(x^2-\frac32x\right)-4 \]
  2. Complete the square

    Complete the square.

    \[ x^2-\frac32x=\left(x-\frac34\right)^2-\frac{9}{16} \]
  3. Simplify surd

    So

    \[ 8x^2-12x-4=8\left(x-\frac34\right)^2-\frac{17}{2} \]
  4. Simplify surd

    Now solve

    \[ 8x^2-12x-4=0 \]
    \[ 8\left(x-\frac34\right)^2-\frac{17}{2}=0 \]
    \[ \left(x-\frac34\right)^2=\frac{17}{16} \]
    \[ x-\frac34=\pm\frac{\sqrt{17}}4 \]
    \[ x=\frac{3\pm\sqrt{17}}4 \]
Final answer

\(8\left(x-\frac34\right)^2-\frac{17}{2}\), and \(x=\frac{3\pm\sqrt{17}}4\).

16. May 2024 P1HR · Q20

Hard5 marks
May 2024 P1HR Question 20 (5 marks): Completing the Square – 20 (a) Express 2 11 9 2 x x in the form a x b c ( ) - - 2 where a, b and c are numbers to be found.  (3) The curve C has equation y x x 2 3 11 3 9 2 ( ) ( ) The point P is the minimum point on C (b) Find the coordinat
Show worked solution
  1. Use Moved Coordinate Geometry Completing

    For part (a),

    \[ 2x^2-11x+9 \]
  2. Factor out 2 from the

    Factor out 2 from the quadratic terms:

    \[ 2\left(x^2-\frac{11}{2}x\right)+9 \]
  3. Complete the square

    Complete the square:

    \[ 2\left[\left(x-\frac{11}{4}\right)^2-\frac{121}{16}\right]+9 \]
    \[ 2\left(x-\frac{11}{4}\right)^2-\frac{121}{8}+9 \]
    \[ 2\left(x-\frac{11}{4}\right)^2-\frac{121}{8}+\frac{72}{8} \]
    \[ 2\left(x-\frac{11}{4}\right)^2-\frac{49}{8} \]
  4. Evaluate fraction

    So

    \[ 2x^2-11x+9=2\left(x-\frac{11}{4}\right)^2-\frac{49}{8} \]
  5. Calculate value

    For part (b),

    \[ y=2(x-3)^2-11(x-3)+9 \]
  6. Let
    \[ u=x-3 \]
  7. Evaluate fraction

    From part (a), the minimum occurs when

    \[ u=\frac{11}{4} \]
  8. Evaluate fraction

    So

    \[ x-3=\frac{11}{4} \]
    \[ x=\frac{23}{4} \]
  9. The minimum value of is

    The minimum value of \(y\) is

    \[ -\frac{49}{8} \]
Final answer

\(\displaystyle 2\left(x-\frac{11}{4}\right)^2-\frac{49}{8}\), \(\displaystyle P=\left(\frac{23}{4},-\frac{49}{8}\right)\)

17. Jun 2022 P1HR · Q23

Hard5 marks
Jun 2022 P1HR Question 23 (5 marks): Completing the Square – 23 (a) Express 2x 2 - 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 (
Show worked solution
  1. (a)
    \[ 2x^2-12x+3=2(x^2-6x)+3 \]
    \[ =2\left((x-3)^2-9\right)+3 \]
    \[ =2(x-3)^2-18+3 \]
    \[ =2(x-3)^2-15 \]
  2. (b)
    \[ y=2(x+4)^2-12(x+4)+3 \]
  3. Let

    Let \(u=x+4\). From part (a),

    \[ 2u^2-12u+3=2(u-3)^2-15 \]
  4. Calculate value

    So

    \[ y=2((x+4)-3)^2-15 \]
    \[ y=2(x+1)^2-15 \]
  5. The minimum point is when

    The minimum point is when \(x=-1\), giving \(y=-15\).

Final answer

(a) \(2(x-3)^2-15\), (b) \(M=(-1,-15)\)

18. Jun 2025 P2H · Q25

Hard6 marks
Jun 2025 P2H Question 25 (6 marks): Completing the Square – 25 (a) Write 28 24 6 2 x x in the form a b x c - - ( )2 where a, b and c are integers. (3) (b) On the axes below, sketch the graph of y x x 28 24 6 2 Show clearly the coordinates of the turning point and the coordinate
Show worked solution
  1. (a)
    \[ 28+24x-6x^2=-6x^2+24x+28 \]
    \[ =-6(x^2-4x)+28 \]
    \[ =-6\left((x-2)^2-4\right)+28 \]
    \[ =52-6(x-2)^2 \]
  2. (b)
    \[ y=52-6(x-2)^2 \]
  3. The graph is a downward

    The graph is a downward-opening parabola.

  4. The turning point is
    \[ (2,52) \]
  5. Calculate value

    The point of intersection with the \(y\)-axis is found by putting \(x=0\):

    \[ y=28 \]
  6. Find intercepts

    So the \(y\)-intercept is

    \[ (0,28) \]
Final answer

(a) \(52-6(x-2)^2\). For the sketch, show a downward parabola with turning point \((2,52)\) and \(y\)-intercept \((0,28)\).

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