AS & A Level · Practice questions · Quadratic equations and inequalities

Roots first, then the region.

Six original Cambridge-style questions on solving quadratic equations and quadratic inequalities, from factorising to a two-condition problem in context.

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

Zero on one side, roots, sketch, read.

01
Recall
[2 marks]

Solve \(2x^{2} - 7x + 3 = 0\).

  • \((2x - 1)(x - 3) = 0\). ✓
  • \(x = \dfrac{1}{2}\) or \(x = 3\). ✓
02
Calculation
[3 marks]

Solve \(5x^{2} - 3x - 1 = 0\), giving your answers correct to 3 significant figures.

  • \(b^{2} - 4ac = 9 + 20 = 29\). ✓
  • \(x = \dfrac{3 \pm \sqrt{29}}{10}\). ✓
  • \(x = 0.839\) or \(x = -0.239\). ✓
03
Calculation
[3 marks]

Solve the inequality \(x^{2} - x - 20 \ge 0\).

  • \((x - 5)(x + 4) = 0\) gives critical values \(5\) and \(-4\). ✓
  • The parabola opens upwards, so \(\ge 0\) outside the roots, ends included. ✓
  • \(x \le -4\) or \(x \ge 5\). ✓
04
Analysis
[3 marks]

Solve the inequality \((2x - 1)(x + 2) < 3x + 4\), giving your answer in exact form.

  • Expand: \(2x^{2} + 3x - 2 < 3x + 4\). ✓
  • Simplify: \(2x^{2} < 6 \Rightarrow x^{2} < 3\). ✓
  • \(-\sqrt{3} < x < \sqrt{3}\). ✓
05
Analysis
[4 marks]

The equation \(kx^{2} + 2kx + 3 = 0\), where \(k\) is a non-zero constant, has no real roots. Find the set of possible values of \(k\).

  • No real roots: \((2k)^{2} - 4(k)(3) < 0\). ✓
  • \(4k^{2} - 12k < 0 \Rightarrow 4k(k - 3) < 0\). ✓
  • Critical values \(0\) and \(3\); a positive quadratic in \(k\) is negative between them. ✓
  • \(0 < k < 3\). ✓
06
Analysis
[5 marks]

A rectangle has length \((x + 5)\) cm and width \((x - 1)\) cm, where \(x > 1\). Its area is at least 40 cm2 and its perimeter is less than 32 cm.

Find the set of possible values of \(x\).

  • Area: \((x + 5)(x - 1) \ge 40 \Rightarrow x^{2} + 4x - 45 \ge 0\). ✓
  • \((x + 9)(x - 5) \ge 0\), so \(x \le -9\) or \(x \ge 5\); with \(x > 1\) this gives \(x \ge 5\). ✓
  • Perimeter: \(2\big[(x + 5) + (x - 1)\big] = 4x + 8 < 32 \Rightarrow x < 6\). ✓
  • Both conditions: \(5 \le x < 6\). ✓

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.