IGCSE · Practice questions · Completing the square

Halve, square, subtract.

Extended. Six original Cambridge-style questions on completing the square, solving equations exactly and finding turning points.

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

Halve \(b\), square it, subtract it.

\(a \ne 1\): take \(a\) out of the \(x^{2}\) and \(x\) terms first.
Halve: \(x^{2} + bx = \left(x + \dfrac{b}{2}\right)^{2} - \left(\dfrac{b}{2}\right)^{2}\).
Tidy: combine the constants; multiply back by \(a\) if you took it out.
Solve: isolate the square, square root with \(\pm\), then rearrange.
Turning point: \(y = a(x + p)^{2} + q\) turns at \((-p,\ q)\).

Recall

[2 marks]

Write in the form \((x + p)^{2} + q\): (a) \(x^{2} + 8x + 3\) (b) \(x^{2} - 2x - 5\).

(a) \((x + 4)^{2} - 16 + 3 = (x + 4)^{2} - 13\) ✓
(b) \((x - 1)^{2} - 1 - 5 = (x - 1)^{2} - 6\) ✓

Calculation

[2 marks]

Write \(x^{2} + 5x - 2\) in the form \((x + p)^{2} + q\).

\(\left(x + \dfrac{5}{2}\right)^{2} - \dfrac{25}{4} - 2\) ✓
\(= \left(x + \dfrac{5}{2}\right)^{2} - \dfrac{33}{4}\) ✓

Calculation

[3 marks]

Write \(3x^{2} - 12x + 5\) in the form \(a(x + p)^{2} + q\).

\(3(x^{2} - 4x) + 5\) ✓
\(= 3[(x - 2)^{2} - 4] + 5\) ✓
\(= 3(x - 2)^{2} - 7\) ✓

Calculation

[4 marks]

By completing the square, solve \(x^{2} + 8x + 3 = 0\).

Give your answers (a) in exact form (b) correct to 2 decimal places.

(a) \((x + 4)^{2} - 13 = 0\), so \((x + 4)^{2} = 13\) ✓
\(x + 4 = \pm\sqrt{13}\), \(x = -4 \pm \sqrt{13}\) ✓
(b) \(x = -0.39\) or \(x = -7.61\) ✓

Analysis

[3 marks]

(a) Write \(5 + 4x - x^{2}\) in the form \(r - (x + p)^{2}\).

(b) Hence write down the maximum value of \(5 + 4x - x^{2}\) and the value of \(x\) where it occurs.

(a) \(-(x^{2} - 4x) + 5 = -[(x - 2)^{2} - 4] + 5 = 9 - (x - 2)^{2}\) ✓
(b) \((x - 2)^{2} \ge 0\), so the expression is largest when \((x - 2)^{2} = 0\) ✓
Maximum value 9, when \(x = 2\) ✓

Proof

[5 marks]

(a) Write \(x^{2} - 4x + 7\) in the form \((x + p)^{2} + q\).

(b) Explain why the equation \(x^{2} - 4x + 7 = 0\) has no solutions.

(a) \((x - 2)^{2} - 4 + 7 = (x - 2)^{2} + 3\) ✓
(b) \((x - 2)^{2} \ge 0\), so \((x - 2)^{2} + 3 \ge 3\). It can never equal 0. ✓
(c) The fraction is greatest when the denominator is least ✓
The least value of \(x^{2} - 4x + 7\) is 3 (when \(x = 2\)) ✓
Greatest value \(= \dfrac{12}{3} = 4\) ✓

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.