IGCSE · Practice questions · Quadratic inequalities

Find the roots, shade the region.

Six original Cambridge-style questions on quadratic inequalities, solved graphically and algebraically, with answers written in the correct form: one interval, or two pieces joined by 'or'.

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

Zero on one side, roots, sketch, read off.

Rearrange so one side is zero; never divide by \(x\). If the \(x^{2}\) term is negative, multiply by \(-1\) and reverse the sign.
Critical values: solve \(f(x) = 0\), usually by factorising.
Sketch the U-shaped curve through the critical values.
Read off: \(< 0\) is between the roots (one interval); \(> 0\) is outside them (two pieces joined by 'or').
Ends: include the critical values only for \(\le\) or \(\ge\). Test one value in the original inequality.

Recall

[2 marks]

Solve \((x + 1)(x - 6) > 0\).

Critical values: \(x = -1\) and \(x = 6\). ✓
U shape; \(> 0\) is above the axis, outside the critical values: \(x < -1\) or \(x > 6\). ✓

Calculation

[3 marks]

Find the set of values of \(x\) for which \(3x^{2} + 10x - 8 \le 0\).

\(3x^{2} + 10x - 8 = (3x - 2)(x + 4)\). ✓
Critical values: \(x = \dfrac{2}{3}\) and \(x = -4\). ✓
\(\le 0\) is on or below the axis, between the critical values with the ends included: \(-4 \le x \le \dfrac{2}{3}\). ✓

Calculation

[3 marks]

Solve the inequality \(x(x - 3) \ge 10\).

Make one side zero: \(x^{2} - 3x - 10 \ge 0\). ✓
\((x - 5)(x + 2) \ge 0\), so the critical values are \(5\) and \(-2\). ✓
Outside the critical values, ends included: \(x \le -2\) or \(x \ge 5\). ✓

Analysis

[3 marks]

Solve \(5 + 4x - x^{2} > 0\).

Multiply by \(-1\) and reverse the sign: \(x^{2} - 4x - 5 < 0\). ✓
\((x - 5)(x + 1) < 0\), so the critical values are \(5\) and \(-1\). ✓
Between the critical values: \(-1 < x < 5\). Check \(x = 0\): \(5 > 0\), true. ✓

Analysis

[4 marks]

(a) Write \(x^{2} - 6x + 13\) in the form \((x - a)^{2} + b\).

(b) Hence explain why \(x^{2} - 6x + 13 > 0\) for all real values of \(x\), and state the solution set of \(x^{2} - 6x + 13 \le 0\).

(a) \(x^{2} - 6x + 13 = (x - 3)^{2} - 9 + 13 = (x - 3)^{2} + 4\). ✓
(b) \((x - 3)^{2} \ge 0\) for every real \(x\), so \((x - 3)^{2} + 4 \ge 4 > 0\). ✓
The curve's minimum point is \((3, 4)\), above the \(x\)-axis, so there are no critical values. ✓
\(x^{2} - 6x + 13 \le 0\) has no real solutions: the solution set is empty. ✓

Analysis

[5 marks]

A stone is thrown upwards so that its height above the ground after \(t\) seconds is \(h = 25t - 5t^{2}\) metres.

(a) Find the set of values of \(t\) for which the stone is more than 30 m above the ground.

(b) For how long is the stone more than 30 m above the ground?

(a) \(25t - 5t^{2} > 30 \Rightarrow 5t^{2} - 25t + 30 < 0 \Rightarrow t^{2} - 5t + 6 < 0\). ✓
\((t - 2)(t - 3) < 0\), so \(2 < t < 3\). ✓
(b) \(3 - 2 = 1\) second. ✓
(c) \(h \ge 35\) would need \(5t^{2} - 25t + 35 \le 0\), that is \(t^{2} - 5t + 7 \le 0\). ✓
\(t^{2} - 5t + 7 = (t - 2.5)^{2} + 0.75 > 0\) for every \(t\), so \(h \ge 35\) is never true: the greatest height is 31.25 m. ✓

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.