Home / IGCSE Add Maths 0606 / Equations, inequalities and graphs / Modulus equations
Cambridge IGCSE · Topic 4.1
IGCSE Add Maths 0606 / Topic 4 / IGCSE

Two cases, then check.

The modulus throws away the sign, so \(|A| = 5\) could mean \(A = 5\) or \(A = -5\). Every modulus equation splits into a positive case and a negative case. Solve both, check each answer against the original equation, and use a graph to see how many solutions to expect.

Topic 4All equations, inequalities and graphs lessons →
The key idea

\(|A| = c\) (with \(c \ge 0\)) means \(A = c\) or \(A = -c\). For \(|ax + b| = cx + d\) solve both cases and reject any root that makes \(cx + d\) negative. For \(|ax + b| = |cx + d|\) either use \(ax + b = \pm(cx + d)\) or square both sides. For \(|ax^{2} + bx + c| = d\) solve \(ax^{2} + bx + c = d\) and \(ax^{2} + bx + c = -d\): up to four solutions. On a graph, each solution is an intersection.

x y O 1 2 3 4 5 2 4 6 1.5 (⅔, 1⅔) (4, 5) y = |2x − 3| y = x + 1 Case +: 2x − 3 = x + 1 x = 4 Case −: 3 − 2x = x + 1 x = ⅔ Two intersections, so two solutions dashed: y = 2x − 3 before reflection
Fig. 1: the graph of y = |2x - 3| (a V with its vertex at x = 1.5) and the line y = x + 1. They meet twice, at (2/3, 5/3) and (4, 5), so |2x - 3| = x + 1 has exactly two solutions. The left arm of the V is the reflected part, where 3 - 2x is used
Section 01

One modulus, two cases.

\(|A|\) is the distance of \(A\) from zero, so it is never negative. If \(|A| = 7\), then \(A\) is 7 units from zero: \(A = 7\) or \(A = -7\). For \(|ax + b| = c\), first make the modulus the subject, then write down both cases and solve each one. If \(c\) is negative there is no solution at all, because a modulus cannot be negative.

Worked example The basic two cases
Solve \(|3x - 2| = 7\).
  1. Case +: \(3x - 2 = 7\), so \(3x = 9\) and \(x = 3\).
  2. Case -: \(3x - 2 = -7\), so \(3x = -5\) and \(x = -\dfrac{5}{3}\).
  3. Check: \(|3(3) - 2| = |7| = 7\) and \(|-5 - 2| = |-7| = 7\).
Answer: \(x = 3\) or \(x = -\frac{5}{3}\)
Worked example Isolate the modulus first
Solve \(2|x + 4| - 1 = 9\).
  1. Add 1 and divide by 2: \(|x + 4| = 5\).
  2. Case +: \(x + 4 = 5\), so \(x = 1\).
  3. Case -: \(x + 4 = -5\), so \(x = -9\).
Answer: \(x = 1\) or \(x = -9\)
Your turn
Solve \(|5 - 2x| = 3\).
Examiner trap

Do not 'remove the bars' and solve only \(ax + b = c\): that finds one answer and loses the other. And never split before isolating: in \(2|x + 4| - 1 = 9\) the case \(2(x + 4) - 1 = -9\) is wrong; the minus belongs to \(|x + 4| = 5\), giving \(x + 4 = -5\).

Section 02

A line on the right: check every root.

In \(|ax + b| = cx + d\) the right-hand side contains \(x\), so it can be negative for some values of \(x\). A modulus is never negative, so any root that makes \(cx + d < 0\) is not a solution. Solve both cases exactly as before, then substitute each root into the ORIGINAL equation. Graphically, you are finding where the line \(y = cx + d\) meets the V-shaped graph \(y = |ax + b|\) (Fig. 1); the line can meet the V twice, once or not at all.

Worked example Two valid roots
Solve \(|2x - 3| = x + 1\).
  1. Case +: \(2x - 3 = x + 1\), so \(x = 4\).
  2. Case -: \(-(2x - 3) = x + 1\), so \(3 - 2x = x + 1\), \(2 = 3x\) and \(x = \dfrac{2}{3}\).
  3. Check \(x = 4\): left \(|5| = 5\), right \(4 + 1 = 5\). Valid.
  4. Check \(x = \dfrac{2}{3}\): left \(\left|\dfrac{4}{3} - 3\right| = \dfrac{5}{3}\), right \(\dfrac{2}{3} + 1 = \dfrac{5}{3}\). Valid.
  5. Fig. 1 shows the line crossing both arms of the V, so two solutions is right.
Answer: \(x = \frac{2}{3}\) or \(x = 4\)
Worked example One root fails the check
Solve \(|x + 5| = 3x - 1\).
  1. Case +: \(x + 5 = 3x - 1\), so \(6 = 2x\) and \(x = 3\).
  2. Case -: \(-(x + 5) = 3x - 1\), so \(-x - 5 = 3x - 1\), \(-4 = 4x\) and \(x = -1\).
  3. Check \(x = 3\): left \(|8| = 8\), right \(9 - 1 = 8\). Valid.
  4. Check \(x = -1\): left \(|4| = 4\), right \(-3 - 1 = -4\). A modulus cannot equal \(-4\), so reject \(x = -1\).
  5. On a graph, the steep line \(y = 3x - 1\) crosses only the right arm of \(y = |x + 5|\).
Answer: \(x = 3\) only
Your turn
Solve \(|x - 4| = 2x + 1\).
Examiner trap

Both cases can give a number without either being a solution, so the check is not optional: it earns the final mark. Check in the ORIGINAL equation, not in the case you solved (a root always satisfies its own case). And it is possible for both roots to fail, giving no solution.

Stage 1 · Learn

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Four quick checks. Each correct answer earns XP and records this skill on your profile.

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Solve \(|x - 2| = 5\).

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How many solutions does \(|2x + 1| = -4\) have?

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Solve \(|x + 1| = 2x - 4\).

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How many solutions does \(|x^{2} - 6x| = 5\) have?

Section 03

Two moduli and quadratic moduli.

For \(|ax + b| = |cx + d|\) both sides are distances, so \(ax + b = \pm(cx + d)\): two linear equations. Alternatively square both sides: \((ax + b)^{2} = (cx + d)^{2}\) is a quadratic with exactly the same solutions, because both sides were already non-negative, so no check is needed. For \(|ax^{2} + bx + c| = d\), the two cases \(ax^{2} + bx + c = d\) and \(ax^{2} + bx + c = -d\) are both quadratics, so there can be up to four solutions. The graph of \(y = |ax^{2} + bx + c|\) (the parabola with its part below the \(x\)-axis reflected up) shows how many: a horizontal line \(y = d\) below the reflected vertex meets it four times.

Worked example Squaring both sides
Solve \(|2x + 1| = |x - 4|\).
  1. Square: \((2x + 1)^{2} = (x - 4)^{2}\), so \(4x^{2} + 4x + 1 = x^{2} - 8x + 16\).
  2. Collect terms: \(3x^{2} + 12x - 15 = 0\), and divide by 3: \(x^{2} + 4x - 5 = 0\).
  3. Factorise: \((x + 5)(x - 1) = 0\), so \(x = -5\) or \(x = 1\).
  4. The \(\pm\) method agrees: \(2x + 1 = x - 4\) gives \(x = -5\); \(2x + 1 = -(x - 4)\) gives \(3x = 3\), \(x = 1\).
  5. Check: \(|-9| = |-9|\) and \(|3| = |-3|\).
Answer: \(x = -5\) or \(x = 1\)
Worked example A quadratic inside the modulus
Solve \(|x^{2} - 4x| = 3\), giving your answers in exact form.
  1. Case +: \(x^{2} - 4x = 3\), so \(x^{2} - 4x - 3 = 0\) and \(x = \dfrac{4 \pm \sqrt{16 + 12}}{2} = \dfrac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}\).
  2. Case -: \(x^{2} - 4x = -3\), so \(x^{2} - 4x + 3 = 0\), \((x - 1)(x - 3) = 0\) and \(x = 1\) or \(x = 3\).
  3. Graph check: \(x^{2} - 4x = (x - 2)^{2} - 4\) has vertex \((2, -4)\), which reflects to \((2, 4)\). The line \(y = 3\) lies below 4, so it meets \(y = |x^{2} - 4x|\) four times.
  4. (With \(d = 4\) there would be three solutions; with \(d > 4\), two.)
Answer: \(x = 2 - \sqrt{7},\ 1,\ 3\) or \(2 + \sqrt{7}\)
Your turn
Solve \(|x - 3| = |2x + 3|\).
Examiner trap

Squaring is safe ONLY when both sides are moduli. Squaring \(|ax + b| = cx + d\) brings back the invalid root, so you still have to check it. When squaring, expand \((x - 4)^{2}\) as \(x^{2} - 8x + 16\), not \(x^{2} + 16\). In \(|x^{2} - 4x| = 3\), forgetting the case \(x^{2} - 4x = -3\) loses two of the four answers.

Stage 2 · Exam

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Solve the equation \(|3x - 4| = |x + 2|\).

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Solve the equation \(|2x + 3| = x + 1\).

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Solve \(|x^{2} - 2x - 8| = 7\), giving your answers in exact form.

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The line \(y = k\) meets the curve \(y = |x^{2} - 4x|\) at exactly three points. Find \(k\).

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Six original Cambridge-style questions
Two-case equations, a line on the right with one root to reject, two moduli by squaring, a quadratic modulus with four exact answers and a sketch that counts solutions.