IGCSE · Practice questions · Modulus equations

Split, solve, check.

Six original Cambridge-style questions on modulus equations: \(|ax + b| = c\), \(|ax + b| = cx + d\) with checking, \(|ax + b| = |cx + d|\) by squaring and \(|ax^{2} + bx + c| = d\) with up to four solutions, plus a graph that counts them.

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

Isolate, split, solve, check.

Isolate the modulus; if it equals a negative number there is no solution.
Split into \(A = B\) and \(A = -B\), bracketing the whole of \(B\).
Two moduli: use \(\pm\) or square both sides; no check is needed.
Check roots of \(|ax + b| = cx + d\) in the original equation; reject any that make \(cx + d < 0\).
Quadratic modulus: solve both quadratics and use the reflected vertex to confirm how many solutions there are.

Recall

[2 marks]

Solve the equation \(|4x + 1| = 9\).

\(4x + 1 = 9\) gives \(x = 2\). ✓
\(4x + 1 = -9\) gives \(4x = -10\), \(x = -\dfrac{5}{2}\). ✓

Calculation

[3 marks]

Solve the equation \(|3x - 2| = x + 6\).

Case +: \(3x - 2 = x + 6\), so \(2x = 8\), \(x = 4\). ✓
Case -: \(-(3x - 2) = x + 6\), so \(2 - 3x = x + 6\), \(-4 = 4x\), \(x = -1\). ✓
Check: \(x = 4\): \(|10| = 10 = 4 + 6\). \(x = -1\): \(|-5| = 5 = -1 + 6\). Both valid. ✓
\(x = -1\) or \(x = 4\). ✓

Calculation

[4 marks]

Solve the equation \(|2x - 1| = 3x + 4\), showing clearly why any root is rejected.

Case +: \(2x - 1 = 3x + 4\), so \(x = -5\). ✓
Check: \(3(-5) + 4 = -11 < 0\), but a modulus cannot be negative, so reject \(x = -5\). ✓
Case -: \(-(2x - 1) = 3x + 4\), so \(1 - 2x = 3x + 4\), \(-3 = 5x\), \(x = -\dfrac{3}{5}\). ✓
Check: left \(\left|-\dfrac{6}{5} - 1\right| = \dfrac{11}{5}\), right \(-\dfrac{9}{5} + 4 = \dfrac{11}{5}\). Valid. ✓
\(x = -\dfrac{3}{5}\) only. ✓

Calculation

[4 marks]

Solve the equation \(|2x - 7| = |x + 1|\).

Square both sides: \(4x^{2} - 28x + 49 = x^{2} + 2x + 1\). ✓
\(3x^{2} - 30x + 48 = 0\), so \(x^{2} - 10x + 16 = 0\). ✓
\((x - 2)(x - 8) = 0\), so \(x = 2\) or \(x = 8\). ✓
Check: \(|-3| = |3|\) and \(|9| = |9|\). ✓

Analysis

[5 marks]

Solve the equation \(|x^{2} - 5x + 1| = 3\), giving your answers in exact form.

Case +: \(x^{2} - 5x + 1 = 3\), so \(x^{2} - 5x - 2 = 0\). ✓
\(x = \dfrac{5 \pm \sqrt{25 + 8}}{2} = \dfrac{5 \pm \sqrt{33}}{2}\). ✓
Case -: \(x^{2} - 5x + 1 = -3\), so \(x^{2} - 5x + 4 = 0\), \((x - 1)(x - 4) = 0\). ✓
\(x = 1\) or \(x = 4\). ✓
So \(x = \dfrac{5 - \sqrt{33}}{2},\ 1,\ 4\) or \(\dfrac{5 + \sqrt{33}}{2}\). (The minimum of \(x^{2} - 5x + 1\) is \(-5.25\), which reflects to \(5.25 > 3\), so four solutions is expected.) ✓

Analysis

[7 marks]

(a) Sketch the graph of \(y = |x^{2} - 2x - 3|\), stating the coordinates of the points where it meets the axes and of the highest point of the reflected section.

(b) Hence state the set of values of \(k\) for which \(|x^{2} - 2x - 3| = k\) has exactly four solutions.

(a) \(x^{2} - 2x - 3 = (x + 1)(x - 3)\), so the curve meets the \(x\)-axis at \((-1, 0)\) and \((3, 0)\), and the \(y\)-axis at \((0, 3)\) (since \(|-3| = 3\)). ✓
\(x^{2} - 2x - 3 = (x - 1)^{2} - 4\): the vertex \((1, -4)\) reflects to the highest point \((1, 4)\) of the arch between \(x = -1\) and \(x = 3\). Outside that interval the curve is the original parabola, rising steeply. ✓
(b) A horizontal line \(y = k\) meets the curve four times when it cuts the arch and both outer arms: \(0 < k < 4\). ✓
(c) Case +: \(x^{2} - 2x - 3 = 3\), so \(x^{2} - 2x - 6 = 0\), \(x = \dfrac{2 \pm \sqrt{28}}{2} = 1 \pm \sqrt{7}\). ✓
Case -: \(x^{2} - 2x - 3 = -3\), so \(x^{2} - 2x = 0\), \(x(x - 2) = 0\), \(x = 0\) or \(x = 2\). ✓
\(x = 1 - \sqrt{7},\ 0,\ 2\) or \(1 + \sqrt{7}\): four solutions, as (b) predicts since \(0 < 3 < 4\). ✓

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.