IGCSE · Practice questions · Solving quadratic equations

Zero on one side, exact on Paper 1.

Six original Cambridge-style questions on solving quadratic equations by factorisation, the formula and completing the square, with exact surd answers and rearranging first.

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, choose a method, simplify, check.

Rearrange to \(ax^{2} + bx + c = 0\): clear fractions (every term) and collect \(x\) terms on one side.
Choose: factorise if you can; complete the square when \(a = 1\) and \(b\) is even; otherwise use the formula.
Exact form: simplify each surd and divide every term of the numerator.
Decimals on Paper 2 only: 3 significant figures unless told otherwise.
Check each root: no zero denominators, no negative lengths, and answer the question asked.

Recall

[3 marks]

Solve (a) \(3x^{2} + 11x - 4 = 0\) (b) \(9x^{2} = 25\).

(a) \(ac = -12\); numbers \(12\) and \(-1\): \(3x(x + 4) - 1(x + 4) = (3x - 1)(x + 4) = 0\). ✓
\(x = \dfrac{1}{3}\) or \(x = -4\). ✓
(b) \(x^{2} = \dfrac{25}{9}\), so \(x = \pm\dfrac{5}{3}\). ✓

Calculation

[3 marks]

By completing the square, solve \(x^{2} + 8x + 3 = 0\), giving your answers in the form \(p \pm \sqrt{q}\).

\(x^{2} + 8x + 3 = (x + 4)^{2} - 16 + 3 = (x + 4)^{2} - 13\). ✓
\((x + 4)^{2} = 13\), so \(x + 4 = \pm\sqrt{13}\). ✓
\(x = -4 \pm \sqrt{13}\). ✓

Calculation

[4 marks]

(a) Without using a calculator, solve \(5x^{2} - 2x - 1 = 0\), giving your answers in the form \(\dfrac{p \pm \sqrt{q}}{r}\).

(b) Give your answers correct to 3 significant figures.

(a) \(b^{2} - 4ac = 4 - 4(5)(-1) = 24\), and \(\sqrt{24} = 2\sqrt{6}\). ✓
\(x = \dfrac{2 \pm 2\sqrt{6}}{10}\). ✓
Divide every term by 2: \(x = \dfrac{1 \pm \sqrt{6}}{5}\). ✓
(b) \(x = 0.690\) or \(x = -0.290\). ✓

Calculation

[4 marks]

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

Multiply every term by \(2(x - 1)(x + 2)\): \(4(x + 2) - 2(x - 1) = (x - 1)(x + 2)\). ✓
\(2x + 10 = x^{2} + x - 2\), so \(x^{2} - x - 12 = 0\). ✓
\((x - 4)(x + 3) = 0\), so \(x = 4\) or \(x = -3\). ✓
Neither makes a denominator zero (\(x \ne 1, -2\)), so both are valid. ✓

Calculation

[3 marks]

Solve \(x(x + 4) = 2(3x + 5)\), giving your answers in the form \(p \pm \sqrt{q}\).

\(x^{2} + 4x = 6x + 10\), so \(x^{2} - 2x - 10 = 0\). ✓
\((x - 1)^{2} - 1 - 10 = 0\), so \((x - 1)^{2} = 11\). ✓
\(x = 1 \pm \sqrt{11}\). ✓

Analysis

[6 marks]

A rectangular lawn measures 10 m by 6 m. A path of constant width \(w\) m is laid all the way round it, so that the lawn and path together cover 100 m².

(a) Show that \(w^{2} + 8w - 10 = 0\).

(b) Find the exact value of \(w\), explaining why only one root is valid.

(a) The outer rectangle is \((10 + 2w)\) m by \((6 + 2w)\) m, so \((10 + 2w)(6 + 2w) = 100\). ✓
\(60 + 32w + 4w^{2} = 100\), so \(4w^{2} + 32w - 40 = 0\); divide by 4: \(w^{2} + 8w - 10 = 0\). ✓
(b) \((w + 4)^{2} - 16 - 10 = 0\), so \((w + 4)^{2} = 26\) and \(w = -4 \pm \sqrt{26}\). ✓
\(-4 - \sqrt{26}\) is negative and a width cannot be negative, so \(w = \sqrt{26} - 4\). ✓
(c) \(w = 1.10\) 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.