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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 Video by Zeeshan Zamurred Functions and Graphs

Edexcel A level Maths: 2.7 Solving Modulus ProblemsWatch the full walkthrough 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 $f (x) = a$ $or$ $f (x) = - a$ , solve both cases. Always $check$ each solution in the original equation, since the modulus can introduce false answers.

∣ f (x) ∣ = a \Rightarrow f (x) = a or f (x) = - a

Two cases — then verify.

1Case 1:

2 x - 1 = 5 \Rightarrow x = 3

2Case 2:

2 x - 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.

∣ 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) ∣ \geq 0 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.

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