Solve modulus equations and reject invalid signed candidates
Use both sign branches, then check original equations when a right-hand expression may be negative.
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Shared Algebra foundations, Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. These modulus and polynomial requirements occur in both sections 2.1 and 3.1. Paper 3 additionally requires partial fractions and rational-power binomial expansions; follow its separate lessons for those methods. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Both sections use linear modulus graphs; nonlinear y = |f(x)| and y = f(|x|) graphs are excluded. Polynomial dividends have degree at most 4 and divisors are linear or quadratic. Written practice is self-checked.
Before you begin
- Interpret modulus as a non-negative value.
- Solve linear equations and substitute candidates into an original equation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For |u| = 7, which signed values can u take?
What you will learn
- Solve |u| = k and |u| = |v| using the necessary sign cases.
- Check a variable right-hand side is non-negative before retaining a squared-equation candidate.
Equal magnitudes allow equal or opposite signed values
For k > 0, |u| = k means u = k or u = −k. For k = 0 it means u = 0, and for k < 0 there is no real solution. For two real expressions, |u| = |v| is equivalent to u = v or u = −v, and also to u² = v². Keep both sign cases; a modulus removes a sign, not one of the possible inputs.
For |3x − 2| = |2x + 7|, the first case 3x − 2 = 2x + 7 gives x = 9. The second case 3x − 2 = −(2x + 7) gives 5x = −5, so x = −1. In the original equation these yield matching magnitudes 25 and 5 respectively. The check confirms both roots rather than relying on one branch.
Squaring needs a sign check when only one side has modulus
For |u(x)| = v(x), the right side must satisfy v(x) ≥ 0. Squaring gives a necessary equation u(x)² = v(x)², but it also allows u(x) = −v(x) when v(x) is negative, which cannot solve the original equation. Solve the branches with their sign restrictions or find candidates and substitute each into the unsquared original. Never assume every squared-equation root is valid.
For |x − 1| = 2x − 5, the right side requires x ≥ 5/2. The branches give candidates x = 4 and x = 2. At x = 4 both sides equal 3; at x = 2 they are 1 and −1. Therefore only x = 4 is valid. A time, length or other contextual domain may impose additional restrictions after the original algebraic equation has been checked.
Pause and explain
Squaring |x − 1| = 2x − 5 gives candidates 2 and 4. Which solves the original equation?
Worked example
Solve |3x − 2| = |2x + 7| and verify each answer.
Show the worked solution
- Use 3x − 2 = 2x + 7 to get x = 9.
- Use 3x − 2 = −(2x + 7) to get x = −1.
- At x = 9 the magnitudes are both 25; at x = −1 they are both 5.
Answer x = −1 or 9; both satisfy the original modulus equation.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Solve |2x − 5| = 7 and check both roots.
Give me a hint
Set the linear inner expression equal to 7 and −7 separately.
Compare my reasoning
- 2x − 5 = 7 gives x = 6.
- 2x − 5 = −7 gives x = −1.
- At these inputs the inner values are 7 and −7, both with magnitude 7.
x = −1 or 6.
Look for these in your work
- Include both signed branches.
- Check each root in the original modulus.
Solve |x − 1| = 2x − 5, explaining any rejected candidate.
Give me a hint
Require a non-negative right side and test both branch solutions.
Compare my reasoning
- The right side must be non-negative, so x ≥ 5/2.
- The branches x − 1 = 2x − 5 and −x + 1 = 2x − 5 give 4 and 2.
- At 4 both sides equal 3; at 2 they are 1 and −1. Reject 2 and retain 4.
x = 4 only; x = 2 is an extraneous signed candidate.
Look for these in your work
- Check the right-side sign in the original equation.
- State why a squared-equation candidate is excluded.
For d(t) = 3t − 12 metres on 0 ≤ t ≤ 6 seconds, find when the distance |d(t)| equals 6 m.
Give me a hint
Use signed separations +6 and −6, then check the time interval.
Compare my reasoning
- 3t − 12 = 6 gives t = 6 s.
- 3t − 12 = −6 gives t = 2 s.
- Both times lie in [0, 6] and give distance 6 m, despite opposite signed separations.
t = 2 s or 6 s.
Look for these in your work
- Solve both distance branches.
- Check the physical time domain and the original magnitude.
Common mistakes
- Include both equal and opposite signed cases.
- A modulus cannot equal a negative right-hand value.
- Verify candidates in the original equation after squaring.
What can you explain now?
Solve |x + 3| = −2 over the reals.
Compare with the explanation
No real solution.
A modulus is non-negative and cannot equal −2.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain when squaring a modulus equation is equivalent and when it produces extra candidates. Check an original right-side sign before accepting a root.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.