2.7PureStretch

Solving Modulus Problems

Solving equations and inequalities with a modulus needs care, because $|f(x)|$ can equal a value in two ways. A sketch plus the two-case approach finds every solution — and helps you discard false ones.

30 min Functions and Graphs

Edexcel A level Maths: 2.7 Solving Modulus ProblemsCurated by Kepler, not produced by us. Watch it before the notes below.

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What you'll be able to do

Solve equations of the form |f(x)| = g(x)
Use the two-case method (±)
Use a sketch to find and check solutions
Solve modulus inequalities

The two cases

Because $|f(x)| = a$ means $f(x) = a$ $\textbf{or}$ $f(x) = -a$ , solve both cases. Always $\textbf{check}$ each solution in the original equation, since the modulus can introduce false answers.

|f(x)| = a \;\Rightarrow\; f(x) = a \ \text{or} \ f(x) = -a

Two cases — then verify.

Example 1Solve

|2x - 1| = 5

1Case 1:

2x - 1 = 5 \Rightarrow x = 3

2Case 2:

2x - 1 = -5 \Rightarrow x = -2

Answer

x = 3

x = -2

Use a sketch

For $|f(x)| = g(x)$ , sketch both $y = |f(x)|$ and $y = g(x)$ ; the solutions are the intersection points. A sketch shows how many solutions to expect and rules out impossible cases (e.g. a modulus can never equal a negative).

Tip — |f(x)| = (a negative number) has NO solutions — the modulus is never negative.

Modulus inequalities

For inequalities like $|f(x)| < a$ , the sketch (or the rule $-a < f(x) < a$ ) gives the solution interval. Reading the regions from a clear sketch is the most reliable method.

Example 1Solve

|x| < 3

|x| < 3

means

-3 < x < 3

Answer

-3 < x < 3

Formula recap

|f(x)| = a \Rightarrow f(x) = \pm a

Two-case method.

|x| < a \Leftrightarrow -a < x < a

Modulus inequality.

|f(x)| \ge 0 \ \text{always}

No solution if RHS < 0.

Common mistakes to avoid

Only solving the positive case.

Solve both f(x) = a and f(x) = −a, then check.

Keeping a solution that fails the original equation.

Always verify — the ± method can produce false solutions.

Key takeaways

|f(x)| = a gives two cases: f(x) = a or f(x) = −a (then verify).
Sketch y = |f(x)| and y = g(x); solutions are the intersections.
|x| < a means −a < x < a; a modulus is never negative.

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

$|2x - 1| = 5$ gives which cases?
- $2x-1 = 5$ or $2x-1 = -5$
- $2x-1 = 5$ only
- $2x = 5$
- $x = 5$
Show answer and explanation
Answer: $2x-1 = 5$ or $2x-1 = -5$
Two cases, ±5.
Solving $|2x - 1| = 5$ gives:
- $x = 3$ or $x = -2$
- $x = 3$ only
- $x = 2$ or $x = -3$
- $x = 5$
Show answer and explanation
Answer: $x = 3$ or $x = -2$
From the two cases.
How many solutions does $|x + 1| = -2$ have?
- None
- One
- Two
- Infinitely many
Show answer and explanation
Answer: None
A modulus is never negative.
$|x| < 3$ means:
- $-3 < x < 3$
- $x > 3$
- $x < 3$ only
- $x < -3$ or $x > 3$
Show answer and explanation
Answer: $-3 < x < 3$
Between −3 and 3.
A reliable way to find all solutions of |f(x)| = g(x) is to:
- Sketch both graphs
- Only solve once
- Square once and stop
- Differentiate
Show answer and explanation
Answer: Sketch both graphs
Intersections are the solutions.
After using the ± method you should always:
- Check solutions in the original
- Double them
- Take the modulus again
- Reject the negative case
Show answer and explanation
Answer: Check solutions in the original
To remove false solutions.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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Related lessons

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Put the methods into practice: see how A-Level maths duels work. An account is required to play; the overview and sample are public.