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Elite Mathematics
IGCSE · 4MA1 · Topic

Solving Quadratics: past paper questions & worked solutions

Every Solving Quadratics question from the 4MA1 papers in our bank, 2020–2025, plus original Elite practice questions, sorted from easiest to hardest. Try each one first, then open the worked solution.

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6questions
20marks in total
6%of papers (3 of 54) include it
0/5/1easy / medium / hard

Questions, easiest first

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

1. May 2025 4WM1H · Q15

Medium3 marks
May 2025 4WM1H Question 15 (3 marks): Solving Quadratics – 15 Solve the equation 5 9 17 0 2 x x Show your working clearly. Give your solutions correct to 3 significant figures. (Total for Question 15 is 3 marks)
Show worked solution
  1. Solve quadratic equation
    \[ 5x^2+9x-17=0 \]
  2. Use quadratic formula

    Use the quadratic formula with \(a=5\), \(b=9\), \(c=-17\).

    \[ x=\frac{-9\pm\sqrt{9^2-4(5)(-17)}}{2(5)} \]
    \[ x=\frac{-9\pm\sqrt{421}}{10} \]
    \[ x=1.152\ldots\quad\text{or}\quad x=-2.952\ldots \]
  3. Solve quadratic equation

    To 3 significant figures,

    \[ x=1.15\quad\text{or}\quad x=-2.95 \]
Final answer

\(x=1.15\) or \(x=-2.95\).

2. Jan 2021 P2HR · Q9

Medium3 marks
Jan 2021 P2HR Question 9 (3 marks): Solving Quadratics – 9 Solve x 2 - 21x + 20 = 0 Show your working clearly. (Total for Question 9 is 3 marks)
Show worked solution
  1. Solve quadratic equation
    \[ x^2-21x+20=0 \]
  2. Factorise

    Factorise: \[ (x-20)(x-1)=0 \]

  3. Solve quadratic equation

    So \[ x=20 \quad \text{or} \quad x=1 \]

Final answer

\(x=20\) or \(x=1\).

3. Jan 2020 P1HR · Q6

Medium3 marks
Jan 2020 P1HR Question 6 (3 marks): Solving Quadratics – 6 Solve x2 - 5x - 36 = 0 Show clear algebraic working. (Total for Question 6 is 3 marks)
Show worked solution
  1. Solve quadratic equation
    \[ x^2-5x-36=0 \]
  2. Factorise

    Factorise: \[ (x-9)(x+4)=0 \]

  3. Solve quadratic equation

    So \[ x=9 \quad \text{or} \quad x=-4 \]

Final answer

\(x=9\) or \(x=-4\).

4. Elite Practice · Q30

Medium3 marks
Elite Practice Question 30 (3 marks): Solving Quadratics – 30 Solve 2x^2 - 5x - 4 = 0. Show your working clearly. Give your solutions correct to 3 significant figures. (Total for Question 30 is 3 marks)
Show worked solution
  1. Identify a, b and c

    \(a=2,\ b=-5,\ c=-4\)

  2. Substitute into the formula
    \[ x=\frac{-(-5)\pm\sqrt{(-5)^2-4(2)(-4)}}{2(2)}=\frac{5\pm\sqrt{57}}{4} \]
  3. Work out both solutions
    \[ x=\frac{5+7.5498\ldots}{4}=3.14,\qquad x=\frac{5-7.5498\ldots}{4}=-0.637 \]
Final answer

\(x=3.14\) or \(x=-0.637\)

5. Elite Practice · Q31

Medium3 marks
Elite Practice Question 31 (3 marks): Solving Quadratics – 31 Solve x^2 + 6x + 2 = 0. Give your solutions in the form a ± sqrt b where a and b are integers. (Total for Question 31 is 3 marks)
Show worked solution
  1. Use the formula
    \[ x=\frac{-6\pm\sqrt{36-8}}{2}=\frac{-6\pm\sqrt{28}}{2} \]
  2. Simplify the surd

    \(\sqrt{28}=2\sqrt7\), so

    \[ x=\frac{-6\pm2\sqrt7}{2}=-3\pm\sqrt7 \]
Final answer

\(x=-3\pm\sqrt7\)

6. Elite Practice · Q32

Hard5 marks
Elite Practice Question 32 (5 marks): Solving Quadratics – 32 A rectangle has length (x + 3) cm, width (x - 1) cm and area 20 cm^2. (a) Show that x^2 + 2x - 23 = 0. (b) Work out the length of the rectangle, correct to 3 significant figures. (Total for Question 32 is 5 marks)
Show worked solution
  1. (a) Area = length × width
    \[ (x+3)(x-1)=20 \;\Rightarrow\; x^2+2x-3=20 \;\Rightarrow\; x^2+2x-23=0 \]
  2. (b) Use the formula
    \[ x=\frac{-2\pm\sqrt{4+92}}{2}=\frac{-2\pm\sqrt{96}}{2}=-1\pm\sqrt{24} \]

    The width \(x-1\) must be positive, so \(x=-1+\sqrt{24}=3.898\ldots\)

  3. (b) Length
    \[ x+3=6.898\ldots\approx6.90\text{ cm} \]
Final answer

(b) \(6.90\) cm

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Dr Eslam Ahmed
Dr Eslam Ahmed

Elite Mathematics · Questions classified by topic, solutions checked by Dr Eslam