IGCSE · Practice questions · Solving quadratic equations

Zero on one side, then solve.

Extended. Six original Cambridge-style questions on solving quadratic equations by factorisation and the quadratic formula.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Rearrange to zero, then factorise or use the formula.

Zero: rearrange to \(ax^{2} + bx + c = 0\).
Factorise: if you can, set each bracket to zero.
Formula: \(x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\); work out \(b^{2} - 4ac\) first.
Form: exact (surds) or rounded, as asked.
Context: reject impossible answers and answer the actual question.

Recall

[3 marks]

Solve (a) \(x^{2} - 5x + 6 = 0\) (b) \(x^{2} + 3x = 10\).

(a) \((x - 2)(x - 3) = 0\), \(x = 2\) or \(x = 3\) ✓
(b) \(x^{2} + 3x - 10 = 0\), \((x + 5)(x - 2) = 0\), \(x = -5\) or \(x = 2\) ✓

Calculation

[3 marks]

Solve \(3x^{2} - 10x + 8 = 0\).

\(3x^{2} - 6x - 4x + 8 = 0\) ✓
\((3x - 4)(x - 2) = 0\) ✓
\(x = \dfrac{4}{3}\) or \(x = 2\) ✓

Calculation

[4 marks]

Solve \(x^{2} - 6x + 4 = 0\).

(a) Give your answers in the form \(p \pm \sqrt{q}\). (b) Give your answers correct to 2 decimal places.

(a) \(b^{2} - 4ac = 36 - 16 = 20\), \(x = \dfrac{6 \pm \sqrt{20}}{2} = \dfrac{6 \pm 2\sqrt{5}}{2}\) ✓
\(x = 3 \pm \sqrt{5}\) ✓
(b) \(x = 5.24\) or \(x = 0.76\) ✓

Calculation

[3 marks]

Solve \(3x^{2} + 7x - 2 = 0\), giving your answers correct to 2 decimal places.

\(b^{2} - 4ac = 49 - 4(3)(-2) = 73\) ✓
\(x = \dfrac{-7 \pm \sqrt{73}}{6}\) ✓
\(x = 0.26\) or \(x = -2.59\) ✓

Calculation

[4 marks]

Solve \(\dfrac{2}{x + 2} + \dfrac{3}{2x - 1} = 1\).

Multiply by \((x + 2)(2x - 1)\): \(2(2x - 1) + 3(x + 2) = (x + 2)(2x - 1)\) ✓
\(7x + 4 = 2x^{2} + 3x - 2\) ✓
\(2x^{2} - 4x - 6 = 0\), so \(x^{2} - 2x - 3 = 0\) ✓
\((x - 3)(x + 1) = 0\), \(x = 3\) or \(x = -1\) ✓

Analysis

[6 marks]

A rectangular garden measures \((x + 4)\) m by \((2x - 1)\) m. Its area is 60 m².

(a) Show that \(2x^{2} + 7x - 64 = 0\).

(b) Solve this equation, giving your answers correct to 2 decimal places.

(a) \((x + 4)(2x - 1) = 2x^{2} - x + 8x - 4 = 2x^{2} + 7x - 4\) ✓
\(2x^{2} + 7x - 4 = 60\), so \(2x^{2} + 7x - 64 = 0\) ✓
(b) \(b^{2} - 4ac = 49 + 512 = 561\), \(x = \dfrac{-7 \pm \sqrt{561}}{4}\) ✓
\(x = 4.17\) or \(x = -7.67\) ✓
(c) \(x = -7.67\) gives negative lengths, so \(x = 4.171...\): sides 8.171... m and 7.342... m ✓
Perimeter \(= 2(8.171... + 7.342...) = 31.0\) m (3 s.f.) ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.