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Solve A-level modulus equations and inequalities using branches, graphs and sign checks. Find valid roots, reject extraneous answers and practise with full solutions.
Before you startThe modulus function, linear inequalities and quadratic equations
01 / A reliable method
An equation |u(x)| = v(x) has a nonnegative left side. At any solution, the right side must therefore be nonnegative too. Use the two possible signs of the argument to generate candidates, then keep only valid ones.
u = v on the branch u ≥ 0
−u = v on the branch u < 0
You may instead solve u = ±v and substitute candidates into the original equation. The substitution check is essential when v depends on x.
02 / Variable right side
Solve |3x − 2| = x + 4. The argument changes sign at x = 2/3. Any solution also needs x + 4 ≥ 0.
For x ≥ 2/3:
3x − 2 = x + 4 ⇒ x = 3
This lies on the correct branch. Original check: |7| = 7.
For x < 2/3:
−3x + 2 = x + 4 ⇒ x = −1/2
This lies on the correct branch. Original check: |−7/2| = 7/2.
Solutions: x = −1/2, 3
The graph of the modulus function meets y = x + 4 twice.
03 / See the intersections
For |2x − 2| = x + k, solving the right-arm equation gives x = k + 2, which is valid only when x ≥ 1. The left-arm equation gives x = (2 − k)/3, which is valid only when x ≤ 1.
For k > −1, both candidates are valid and distinct. At k = −1 they both give the same vertex x = 1. For k < −1 neither lies on its intended arm, so there are no solutions. Move the line to see each case.
The equality sign at the boundary is allowed in both arm formulas for this calculation, but the vertex still counts as only one intersection.
Green V: y = |2x − 2|. Gold line: y = x + k. Blue points mark solutions of the original equation.
k = 1: two solutions, x = 1/3 and x = 3.
Both branch candidates lie on the branches used to obtain them.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 / Reject false candidates
Consider |2x + 5| = x + 1. Solving the two signed equations produces x = −4 and x = −2. Neither is a solution to the original modulus equation.
2x + 5 = x + 1 ⇒ x = −4
Original check: |−3| = 3, but x + 1 = −3.
−(2x + 5) = x + 1 ⇒ x = −2
Original check: |1| = 1, but x + 1 = −1.
No real solutions
Both candidates have a negative right side. Equivalently, each candidate violates the sign branch that produced it.
05 / Between or outside
To solve |2x − 3| < x + 3, the right side must be positive. Under that condition, the argument must lie strictly between the two bounds:
−(x + 3) < 2x − 3 < x + 3
The left inequality gives x > 0. The right gives x < 6. Together these give 0 < x < 6, and throughout this interval x + 3 > 0 as required. The V is below the line between its intersections.
|2x − 3| < x + 3 ⇔ 0 < x < 6
For the complementary inequality |2x − 3| ≥ x + 3, the answer is x ≤ 0 or x ≥ 6. The endpoints are included because equality is now allowed. A test input in each region, such as −1, 2 and 7, confirms the direction.
06 / Sign and squaring traps
For |u| < v, a nonpositive v makes the inequality impossible. For |u| > v, a negative v makes it automatically true. When v = 0, |u| > 0 holds precisely when u ≠ 0.
When v > 0, |u| < v means −v < u < v, while |u| > v means u < −v or u > v. Handle each sign region if v depends on x.
Squaring |u| = v gives u² = v², but loses the information v ≥ 0. Keep that condition or check every candidate afterwards.
|x − 2| = x − 4
requires x ≥ 4
The right side must be nonnegative.
(x − 2)² = (x − 4)²
⇒ 4x = 12 ⇒ x = 3
This candidate violates x ≥ 4.
At x = 3: 1 ≠ −1
Therefore the original equation has no solution.
07 / A quadratic argument
For |x² − 4| = x + 2, the argument changes sign at x = −2 and x = 2. The right side requires x ≥ −2.
Nonnegative argument:
x² − 4 = x + 2
⇒ (x − 3)(x + 2) = 0
Candidates 3 and −2 both have x² − 4 ≥ 0 and satisfy the original equation.
Negative argument:
4 − x² = x + 2
⇒ (x + 2)(x − 1) = 0
x = 1 lies inside (−2, 2) and is valid. The other candidate −2 is a boundary already counted.
Solutions: x = −2, 1, 3
Original outputs are 0, 3 and 5 respectively, matching the right side.
08 / Your turn
For inequalities, write intervals or a clear “and/or” statement. For equations, substitute every proposed answer into the original expression.
Solve |2x − 1| = x + 2.
The two candidates come from 2x − 1 = ±(x + 2).
The positive-argument equation gives x = 3; the negative-argument equation gives x = −1/3. Both are on the correct side of 1/2 and satisfy the original equation. Answers: −1/3 and 3.
Solve |3x + 6| = x + 1.
Check whether x + 1 is nonnegative at your candidates.
3x + 6 = x + 1 gives x = −5/2; −3x − 6 = x + 1 gives x = −7/4. In both cases x + 1 is negative. Neither works, so there are no real solutions.
Solve |3x − 1| ≤ 5.
Use −5 ≤ 3x − 1 ≤ 5.
Adding 1 gives −4 ≤ 3x ≤ 6, hence −4/3 ≤ x ≤ 2. Both endpoints are included.
Solve |2x + 3| > 7.
The argument must be less than −7 or greater than 7.
2x + 3 < −7 gives x < −5. Alternatively, 2x + 3 > 7 gives x > 2. Answer: x < −5 or x > 2.
Solve |x + 2| < 4 − x.
The right side must be positive; then solve −(4 − x) < x + 2 < 4 − x.
The left inequality reduces to −4 < 2, always true. The right gives 2x < 2, so x < 1. This automatically satisfies 4 − x > 0. The full answer is x < 1.
Solve (a) |x − 7| > −2, (b) |x − 7| ≤ −2 and (c) |x − 7| > 0.
Modulus is nonnegative.
(a) Every real x. (b) No real x. (c) Every real x except 7, because only x = 7 makes the modulus zero.
Solve |x² − 1| = 2.
Consider x² − 1 = 2 and x² − 1 = −2.
The first gives x² = 3, so x = ±√3. The second gives x² = −1, which has no real roots. Both surviving roots give modulus 2.
For the model equation |2x − 2| = x + k, find k when there is exactly one real solution, and state the solution.
The two valid intersections coincide at the vertex.
The vertex is (1, 0). For the line to pass through it, 0 = 1 + k, so k = −1. Both branch calculations give x = 1, which is one distinct solution. For k > −1 there are two; for k < −1 there are none.
09 / Recap
Section 1 of 9 · A reliable method