AS & A Level · Practice questions · Completing the square

Square it, then read it.

Six original Cambridge-style questions on completing the square: rewriting quadratics, finding vertices, ranges and greatest values, solving exactly and proving a quadratic is always positive.

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

The method

Halve, square, subtract, tidy.

Factor \(a\) out of the \(x^{2}\) and \(x\) terms if \(a \ne 1\).
Halve the coefficient of \(x\) inside the bracket: \(x^{2} + bx = \left(x + \frac{b}{2}\right)^{2} - \frac{b^{2}}{4}\).
Multiply back by \(a\), including the subtracted square, then collect constants.
Read off \(a(x + p)^{2} + q\): vertex \((-p, q)\), minimum if \(a > 0\), maximum if \(a < 0\).
Check by expanding: you must get the original quadratic back.

Recall

[2 marks]

Express \(x^{2} - 14x + 50\) in the form \((x - a)^{2} + b\).

Half of \(-14\) is \(-7\): \((x - 7)^{2} - 49 + 50\). ✓
\(= (x - 7)^{2} + 1\), so \(a = 7,\ b = 1\). ✓

Calculation

[3 marks]

(a) Express \(3x^{2} - 18x + 32\) in the form \(a(x + b)^{2} + c\).

(b) State the coordinates of the vertex of the curve \(y = 3x^{2} - 18x + 32\).

(a) \(3(x^{2} - 6x) + 32 = 3\big[(x - 3)^{2} - 9\big] + 32\). ✓
\(= 3(x - 3)^{2} - 27 + 32 = 3(x - 3)^{2} + 5\). ✓
(b) The bracket is zero at \(x = 3\), so the vertex is \((3,\ 5)\), a minimum. ✓

Calculation

[3 marks]

By completing the square, solve \(x^{2} + 4x - 8 = 0\), giving your answers in the form \(p \pm q\sqrt{3}\).

\(x^{2} + 4x - 8 = (x + 2)^{2} - 4 - 8 = (x + 2)^{2} - 12\). ✓
\((x + 2)^{2} = 12 \Rightarrow x + 2 = \pm\sqrt{12} = \pm 2\sqrt{3}\). ✓
\(x = -2 \pm 2\sqrt{3}\). ✓

Analysis

[4 marks]

The function f is defined by \(f(x) = 5 + 8x - 2x^{2}\) for \(x \in \mathbb{R}\).

(a) Express \(f(x)\) in the form \(a - b(x - c)^{2}\).

(b) State the range of f and the coordinates of the stationary point of \(y = f(x)\).

(a) \(f(x) = 5 - 2(x^{2} - 4x) = 5 - 2\big[(x - 2)^{2} - 4\big]\). ✓
\(= 5 + 8 - 2(x - 2)^{2} = 13 - 2(x - 2)^{2}\). ✓
(b) \(2(x - 2)^{2} \ge 0\), so \(f(x) \le 13\). Range: \(f(x) \le 13\). ✓
The stationary point is the maximum point \((2,\ 13)\). ✓

Proof

[3 marks]

(a) Show that \(x^{2} - 6x + 13 > 0\) for all real values of \(x\).

(b) Hence find the greatest value of \(\dfrac{1}{x^{2} - 6x + 13}\).

(a) \(x^{2} - 6x + 13 = (x - 3)^{2} - 9 + 13 = (x - 3)^{2} + 4\). ✓
\((x - 3)^{2} \ge 0\) for all real \(x\), so \((x - 3)^{2} + 4 \ge 4 > 0\). ✓
(b) The denominator is least, 4, when \(x = 3\), so the fraction is greatest there: \(\dfrac{1}{4}\). ✓

Analysis

[5 marks]

A ball is thrown upwards. Its height above the ground \(t\) seconds later is \(h = 1.5 + 14t - 5t^{2}\) metres.

(a) Express \(h\) in the form \(a - b(t - c)^{2}\).

(b) Find the greatest height of the ball and the time at which it is reached.

(a) \(h = 1.5 - 5(t^{2} - 2.8t) = 1.5 - 5\big[(t - 1.4)^{2} - 1.96\big]\). ✓
\(= 1.5 + 9.8 - 5(t - 1.4)^{2} = 11.3 - 5(t - 1.4)^{2}\). ✓
(b) Greatest height \(11.3\) m, when \(t = 1.4\) s. ✓
(c) \(h = 0\): \(5(t - 1.4)^{2} = 11.3 \Rightarrow (t - 1.4)^{2} = 2.26\). ✓
\(t - 1.4 = \pm\sqrt{2.26} = \pm 1.503...\); the time must be positive, so \(t = 1.4 + 1.503... = 2.90\) s. ✓

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.