A-level Maths revision · Pure, year 2

Modulus equations and inequalities

The idea

∣x∣|x| is the distance from zero, so it is never negative. ∣f(x)∣=k|f(x)| = k means f(x)=kf(x) = k or f(x)=−kf(x) = -k. ∣f(x)∣<k|f(x)| < k means −k<f(x)<k-k < f(x) < k.

  1. Split into the two cases: positive and negative inside.
  2. Solve each case.
  3. Check each answer in the original equation: some may not work.

Watch out When the right-hand side contains xx, check every solution: one may be false.

See it worked

Worked example

Solve ∣2x−3∣=5|2x - 3| = 5.

  1. 2x−3=52x - 3 = 5 gives x=4x = 4.
  2. 2x−3=−52x - 3 = -5 gives x=−1x = -1.
  3. Both check: ∣5∣=5|5| = 5 and ∣−5∣=5|-5| = 5.

Answer x=4x = 4 or x=−1x = -1

Your turn

Question 1Non-calculator3 marks

Solve ∣x+2∣=7|x + 2| = 7.

Question 2Non-calculator3 marks

Solve ∣3x−1∣<8|3x - 1| < 8.

Question 3Non-calculator3 marks

Solve ∣2x−1∣=x+4|2x - 1| = x + 4.

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