Inequalities with the modulus sign

Sketch first

Draw y = |f(x)| by reflecting the parts of y = f(x) below the x-axis. Draw the other side (often a straight line) on the same axes.

The sketch tells you how many intersections there are and which piece of the modulus graph each one is on.

Find the crossing points

On a piece where f(x) ≥ 0, solve f(x) = g(x). On a reflected piece, solve -f(x) = g(x).

Reject any root that is not on the piece you used (or that makes a modulus equal to a negative number). In June 2025 some students introduced extra critical values, such as ±4, by cross-multiplying incorrectly, and some kept a value that was not admissible (examiners' report): check each one against the sketch.

Squaring

If both sides are non-negative, for example |x - 1| > |2x + 1|, you may square: (x - 1)² > (2x + 1)². This gives a quadratic inequality with no cases to check.

Do not square when one side can be negative, such as |x + 2| ≥ 2x - 1.

Reading off the answer

Use the sketch: where is the modulus graph below (for <) or higher than (for >) the other graph? Write all intervals, and decide whether each end is included (strict or not).

Worked example

Solve |x² - x - 6| < 2 - x.

1. x² - x - 6 = (x - 3)(x + 2), so the graph is reflected for -2 < x < 3.
2. Unreflected piece: x² - x - 6 = 2 - x gives x² = 8, so x = ±2√2. Only x = -2√2 lies on that piece (x ≤ -2).
3. Reflected piece: -(x² - x - 6) = 2 - x gives x² - 2x - 4 = 0, so x = 1 ± √5. Only x = 1 - √5 lies in -2 < x < 3.
4. From the sketch the curve is lower than the line between these two crossings.

Answer: -2√2 < x < 1 - √5

Check yourself

1. Solve |2x - 3| < 5.
   • x < -1 or x > 4
   • -4 < x < 1
   • -1 < x < 4
   • x < 4
   Show answer

   -1 < x < 4. -5 < 2x - 3 < 5, so -2 < 2x < 8, so -1 < x < 4.

2. Solve |x - 1| > |2x + 1|.
   • -2 < x < 0
   • x < -2 or x > 0
   • x > 0
   • -1/2 < x < 1
   Show answer

   -2 < x < 0. Both sides are non-negative, so square: x² - 2x + 1 > 4x² + 4x + 1, so 0 > 3x² + 6x = 3x(x + 2). This holds for -2 < x < 0.

3. The solution of |x² - 4| < 3x is a < x < b. Find b.
   Show answer

   4. For x ≥ 2: x² - 4 = 3x gives x = 4 (x = -1 is rejected). For x < 2: 4 - x² = 3x gives x = 1 (x = -4 rejected because 3x < 0 there). The curve is lower than the line for 1 < x < 4.

4. Using the same inequality |x² - 4| < 3x, find a.
   Show answer

   1. On the reflected piece (-2 < x < 2): 4 - x² = 3x, so x² + 3x - 4 = 0, (x + 4)(x - 1) = 0. Only x = 1 lies on this piece with 3x positive.

5. Find the largest value of x for which |x + 2| ≥ 2x - 1.
   Show answer

   3. x + 2 = 2x - 1 gives x = 3. -(x + 2) = 2x - 1 gives x = -1/3, but there 2x - 1 < 0, so that is not a crossing. The line is lower than the V for all x ≤ 3.

6. Solve 3/|x + 1| > 1.
   • -4 < x < 2
   • -4 < x < 2, x ≠ -1
   • x < -4 or x > 2
   • x > 2
   Show answer

   -4 < x < 2, x ≠ -1. |x + 1| is positive (x ≠ -1), so multiply: |x + 1| < 3, so -4 < x < 2. x = -1 is excluded because the fraction is undefined.

7. Solve |x² - 5x| = 6.
   Show answer

   x = -1, 2, 3 or 6. x² - 5x = 6 gives (x - 6)(x + 1) = 0. x² - 5x = -6 gives (x - 2)(x - 3) = 0. All four satisfy the original equation.

8. When is it valid to square each side of an inequality P < Q?
   • Always
   • Only when P and Q are each negative
   • When P and Q are each known to be non-negative
   • Only when the inequality is strict
   Show answer

   When P and Q are each known to be non-negative. For non-negative numbers, P < Q exactly when P² < Q². If P could be negative this fails: -3 < 2 but 9 > 4.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit F2
• Pearson Edexcel IAL Further Pure Mathematics 2 scheme of work
• Pearson Edexcel IAL Further Pure Mathematics F2 (WFM02) examiners' report, June 2025

Last updated 8 November 2026. Spotted a mistake? Tell us.