Distance, not direction
Modulus — |x| = x when x ⩾ 0, and |x| = −x when x < 0 — the distance of x from zero, so never negative.
The modulus of a number is its size with the sign discarded: |5| = 5 and |−5| = 5. A cleaner way to think of it is as a distance from zero, because distance is never negative, and that reading is what makes every later result obvious rather than something to memorise.
The consequence that causes all the trouble is that two different inputs give the same output. So an equation containing a modulus generally has two solutions, and losing one of them is the standard error in this topic.
- |x|
- the modulus of xits distance from zero
- |x − a|
- distance from x to athe interpretation to reach for first
- √(x²)
- always the positive rootwhich is why it equals |x| and not x
The graph is a fold
To sketch y = |f(x)|, draw y = f(x) first and then reflect everything below the x-axis upwards. Nothing above the axis moves, because it is already positive.
For a linear expression this produces a V. The corner sits where the bracket equals zero — for y = |2x − 6| that is at x = 3 — and the two arms have gradients that are negatives of each other.
On The fold, the dashed line is the original and the solid V is its modulus. Then switch to the equation and inequality and drag k: notice the two intersection points, and that the inequality gives one interval between them rather than two separate regions.
Only linear insides are required
The syllabus asks for graphs of y = |ax + b|. Graphs of y = |f(x)| and y = f(|x|) for non-linear f are explicitly excluded, so you will not be asked to fold a cubic. Knowing the boundary saves revision time.
Solving equations: square, or split into cases
There are two reliable methods and both are accepted. Squaring uses |a| = |b| ⟺ a² = b², which removes the modulus entirely and leaves an ordinary quadratic. It is quick and safe when both sides are moduli.
The case method removes the bars by considering each possibility separately: ax + b = k and ax + b = −k. It generalises better and makes it obvious why two answers appear.
Solve |3x − 2| = |2x + 7|.
- Both sides are moduli, so square them: (3x − 2)² = (2x + 7)².Squaring is valid here because |a| = |b| exactly when a² = b². No extra solutions can be introduced by this step.
- Expand: 9x² − 12x + 4 = 4x² + 28x + 49.Ordinary expansion, with care over the middle terms.
- Rearrange: 5x² − 40x − 45 = 0, so x² − 8x − 9 = 0.Dividing through by 5 keeps the numbers small.
- Factorise: (x − 9)(x + 1) = 0, giving x = 9 or x = −1.Two solutions, as expected from an equation with moduli on both sides.
- Check: |25| = |25| ✓ and |−5| = |5| ✓.Squaring can introduce false solutions when only one side is a modulus, so checking is a good habit.
x = 9 or x = −1
Squaring one side only can invent solutions
For an equation like |x − 1| = 2x, squaring produces a quadratic whose roots include values making the right-hand side negative — impossible, since a modulus cannot equal a negative number. Always substitute your answers back when only one side carried a modulus, and discard any that fail.
Inequalities become intervals
Reading |x − a| as a distance turns every modulus inequality into a sentence about how far x may stray from a, and the answer is then immediate.
|x − a| < b says x is within b of a, which is the interval a − b < x < a + b. The reverse, |x − a| > b, says x is further than b away, giving two separate pieces: x < a − b or x > a + b.
- a
- the centrethe value the bracket is measured from
- b
- the allowed distancemust be positive for solutions to exist
The results worth carrying into the exam
|x| ⩾ 0always, so|f(x)| = negativehas no solutions.- Squaring is safe when both sides are moduli, and needs a check when only one is.
- The corner of
y = |ax + b|is atx = −b/a. - Less-than inequalities give a single interval; greater-than gives two separate regions.
- Sketching the two graphs and reading off where one is above the other works for every case and is often the fastest route.
Reading a modulus graph off the page
Many questions give the graph rather than the equation and ask you to work backwards. Three features fix it completely, and reading them in order is quicker than trying to fit an equation by inspection.
The corner sits where the bracket is zero, so its x-coordinate gives the constant: a corner at x = 4 means the bracket is x − 4. The gradient of the right-hand arm gives the coefficient, since the arms have gradients ±a. And if the whole V has been lifted off the x-axis, a constant has been added outside the modulus.
| Feature of the graph | What it tells you |
|---|---|
| corner at x = k | the bracket is (x − k) |
| right arm has gradient a | the coefficient of x is a |
| corner sits on the x-axis | nothing is added outside |
| corner lifted to y = c | the equation ends "+ c" |
| V opens downwards | there is a minus in front of the modulus |
Inside the bars or outside?
A constant inside the modulus slides the corner sideways: y = |x − 3| has its corner at x = 3. A constant outside lifts the whole graph: y = |x| + 3 keeps its corner at x = 0 but raises it to y = 3. Deciding which one a graph shows is simply a question of whether the corner moved along the axis or off it.