Modulus Equations and Inequalities

Solve equations and inequalities with a modulus by splitting into two cases or squaring, check every answer, and read solutions from a graph.

In the exam for IGCSE Additional Maths 0606 · 14 real past-paper questions below

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One modulus, two cases

A thermostat is set to 21°. It switches on when the room is more than 2° away from 21°, too hot or too cold. “Distance from 21” is a modulus: $∣ T - 21∣$ .

A modulus hides a sign. If $∣ A ∣ = 5$ , the thing inside was either 5 or −5. So every modulus equation splits into two ordinary equations.

Formula

∣ A ∣ = c ⟺ A = c or A = - c

$A$ — the expression inside the modulus, for example $3 x - 4$ .
$c$ — a number that is 0 or more. A modulus can never equal a negative number.

Worked example

1
Solve $∣3 x - 4∣ = 11$ .
2
$3 x - 4 = 11 or 3 x - 4 = - 11$ — Write both cases. This line earns the method mark.
3
$3 x = 15 \Rightarrow x = 5$ — Solve the first case.
4
$3 x = - 7 \Rightarrow x = - \frac{7}{3}$ — Solve the second case. Keep the exact fraction.
5
$∣15 - 4∣ = 11, ∣ - 7 - 4∣ = 11$ — Check both in the original equation. Both work.

Common mistake

Solving only $3 x - 4 = 11$ . That finds one answer and loses the other. Always write the minus case too.

Common mistake

Trying to solve $∣2 x + 1∣ = - 3$ . A modulus is never negative, so there are no solutions.

Rule

On Paper 1 there is no calculator. Leave answers like $- \frac{7}{3}$ as exact fractions.

Key idea

One modulus, two cases. Write both cases down every time.

Modulus equals a line

Now the right-hand side has x in it, for example $∣ x - 6∣ = 2 x$ . You still split into two cases. But there is a trap.

The left side is a modulus, so it is never negative. The right side must therefore be 0 or more too. A case that makes the right side negative gives a false answer.

Worked example

1
Solve $∣ x - 6∣ = 2 x$ .
2
$x - 6 = 2 x \Rightarrow x = - 6$ — First case: the inside equals the right side.
3
$2 x = 2 (- 6) = - 12 < 0$ — Check: the right side would be negative, but a modulus cannot equal −12. Reject x = −6.
4
$x - 6 = - 2 x \Rightarrow 3 x = 6 \Rightarrow x = 2$ — Second case: the inside equals minus the right side.
5
$∣2 - 6∣ = 4 = 2 (2)$ — Check: both sides are 4. So x = 2 is the only solution.

The graph shows why. The line $y = 2 x$ meets the V of $y = ∣ x - 6∣$ only once, on its left arm.

Common mistake

Keeping every answer from the two cases. With x on the right-hand side, put each answer back into the original equation and throw away any that fail.

Key idea

When the right side contains x, solve both cases, then check both. One of them is often false.

Modulus on both sides

When there is a modulus on each side, both sides are sizes. Either the insides are equal, or one is minus the other.

Formula

∣ A ∣ = ∣ B ∣ ⟺ A = B or A = - B

$A$ — the expression inside the left modulus.
$B$ — the expression inside the right modulus.

Worked example

1
Solve $∣2 x - 3∣ = ∣ x + 6∣$ .
2
$2 x - 3 = x + 6 \Rightarrow x = 9$ — First case: the insides are equal.
3
$2 x - 3 = - (x + 6) \Rightarrow 3 x = - 3 \Rightarrow x = - 1$ — Second case: put a bracket round the whole right side before the minus.

There is a second method. Both sides are 0 or more, so you may square them. Squaring removes the modulus signs.

Worked example

1
Solve $∣2 x - 3∣ = ∣ x + 6∣$ by squaring.
2
$(2 x - 3)^{2} = (x + 6)^{2}$ — Square both sides. The modulus signs disappear.
3
$4 x^{2} - 12 x + 9 = x^{2} + 12 x + 36$ — Expand both brackets.
4
$3 x^{2} - 24 x - 27 = 0$ — Collect everything on one side.
5
$x^{2} - 8 x - 9 = (x - 9) (x + 1) = 0$ — Divide by 3 and factorise.
6
$x = 9, x = - 1$ — The same answers as before.

Common mistake

Writing $2 x - 3 = - x + 6$ . The minus must reach every term: $- (x + 6) = - x - 6$ .

Key idea

Modulus on both sides: use the two cases, or square both sides. Both methods earn full marks.

A quadratic inside

Now the inside is a quadratic, for example $∣ x^{2} - 3 x - 4∣ = 6$ . Split into two cases as usual. This time each case is a quadratic equation.

Worked example

1
Solve $∣ x^{2} - 3 x - 4∣ = 6$ .
2
$x^{2} - 3 x - 4 = 6$ — First case: the inside equals 6.
3
$x^{2} - 3 x - 10 = (x - 5) (x + 2)$ $= 0$ $\Rightarrow x = 5, - 2$ — Rearrange to equal 0 and factorise.
4
$x^{2} - 3 x - 4 = - 6$ — Second case: the inside equals −6.
5
$x^{2} - 3 x + 2 = (x - 1) (x - 2) = 0$ $\Rightarrow x = 1, 2$ — Rearrange and factorise again.
6
$x = - 2, 1, 2, 5$ — Four solutions. The line y = 6 cuts both outer arms and both sides of the hump.

Sometimes the second case has no real roots. Then the line $y = d$ is above the hump and misses it.

Worked example

1
Solve $∣ x^{2} - 3 x - 4∣ = 8$ .
2
$x^{2} - 3 x - 12 = 0 \Rightarrow x = \frac{3 \pm 57}{2}$ — First case. It does not factorise, so use the quadratic formula. Keep the surd exact.
3
$x^{2} - 3 x + 4 = 0 : b^{2} - 4 a c = 9 - 16 = - 7$ — Second case. The discriminant is negative, so this case gives no solutions.
4
$x = \frac{3 \pm 57}{2}$ — Only two solutions.

Common mistake

Solving only the positive case. For $∣ x^{2} - 3 x - 4∣ = 6$ that finds 2 of the 4 answers, and loses half the marks.

Key idea

Quadratic inside: two cases, so up to four answers. A negative discriminant means that case adds none.

Inequalities with a number

Inequalities follow the same picture. $∣ A ∣ < c$ means A is less than c away from 0. $∣ A ∣ > c$ means A is more than c away from 0.

Formula

∣ A ∣ < c ⟺ - c < A < c

$A$ — the expression inside the modulus.
$c$ — a positive number. The answer is one piece, sandwiched between −c and c.

Formula

∣ A ∣ > c ⟺ A < - c or A > c

$A$ — the expression inside the modulus.
$c$ — a number that is 0 or more. The answer is two separate pieces.

The same is true with $\leq$ and $\geq$ .

Worked example

1
Solve $∣2 x - 3∣ \leq 7$ .
2
$- 7 \leq 2 x - 3 \leq 7$ — Less than: write one sandwich.
3
$- 4 \leq 2 x \leq 10$ — Add 3 to all three parts.
4
$- 2 \leq x \leq 5$ — Divide all three parts by 2.

Worked example

1
Solve $3∣ x + 2∣ > 6$ .
2
$∣ x + 2∣ > 2$ — Divide by 3 first, so the modulus is on its own.
3
$x + 2 < - 2 or x + 2 > 2$ — Greater than: two separate pieces.
4
$x < - 4 or x > 0$ — Subtract 2 in each piece.

Memory trick

Less than — the answer is trapped in between. Greater than — the answer escapes to both sides.

Common mistake

Writing $- 4 > x > 0$ . No number is below −4 and above 0 at the same time. Two pieces must be written with “or”.

Key idea

The critical values come from the equation. The type of inequality decides whether the answer is between them or outside them.

Harder inequalities

For $k ∣ a x + b ∣ \leq ∣ c x + d ∣$ , both sides are 0 or more. So you can square both sides safely. You get an ordinary quadratic inequality.

Worked example

1
Solve $2∣ x - 1∣ \leq ∣ x + 4∣$ .
2
$4 (x - 1)^{2} \leq (x + 4)^{2}$ — Square both sides. The 2 becomes 4.
3
$4 x^{2} - 8 x + 4 \leq x^{2} + 8 x + 16$ — Expand.
4
$3 x^{2} - 16 x - 12 \leq 0$ — Collect on one side.
5
$(3 x + 2) (x - 6) \leq 0$ — Factorise. The critical values are $- \frac{2}{3}$ and 6.
6
$- \frac{2}{3} \leq x \leq 6$ — A positive quadratic is below 0 between its roots.

For $∣ a x + b ∣ \leq c x + d$ , you must not square, because the right side can be negative. Find the critical values from the equation, check them, then use a sketch.

Worked example

1
Solve $∣ x - 3∣ \leq x + 1$ .
2
$x - 3 = x + 1$ — First case: the x terms cancel and leave −3 = 1, which is impossible. No solution here.
3
$x - 3 = - (x + 1) \Rightarrow x = 1$ — Second case gives the only critical value.
4
$x = 5 : ∣2∣ = 2 \leq 6$ — Test a point to the right of 1: it works.
5
$x = 0 : ∣ - 3∣ = 3 > 1$ — Test a point to the left of 1: it fails. The V is above the line there.
6
$x \geq 1$ — The final answer.

Common mistake

Squaring when one side is not a modulus. If that side can be negative, squaring creates false answers.

Rule

Mark schemes give a method mark for the two cases (or for squaring), accuracy marks for the critical values, and a final mark for the answer in the correct form.

Key idea

Modulus on both sides: square. A line on one side: find the critical values, then test or sketch.

Quadratic inequalities by graph

For a quadratic inside a modulus, such as $∣ x^{2} - 3 x - 4∣ \leq 6$ , use the graph. You already know the critical values from the equation: $- 2, 1, 2, 5$ .

Steps

1
1. Sketch $y = ∣ x^{2} - 3 x - 4∣$ : cusps at $- 1$ and $4$ , a hump of height $\frac{25}{4}$ at $x = \frac{3}{2}$ .
2. Draw the flat line $y = 6$ . It meets the graph at the four critical values.
3. For $\leq 6$ , choose where the graph is on or below the line. For $> 6$ , choose where it is above.

Worked example

1
Solve $∣ x^{2} - 3 x - 4∣ \leq 6$ .
2
$- 2, 1, 2, 5$ — The critical values, from the two cases.
3
$- 2 \leq x \leq 1$ — The graph is under the line from the left arm up to the left side of the hump.
4
$2 \leq x \leq 5$ — It is under the line again from the right side of the hump to the right arm.
5
$- 2 \leq x \leq 1 or 2 \leq x \leq 5$ — Between 1 and 2 the hump pokes above the line, so those x values are left out.

For $∣ x^{2} - 3 x - 4∣ > 6$ you take the other parts: $x < - 2$ , $1 < x < 2$ or $x > 5$ .

Rule

If the question gives a grid and asks you to draw the graph, draw it accurately. Then read off the crossing points and write the answer as inequalities.

Common mistake

Forgetting the middle piece. The hump can poke above the line, which splits the answer into two parts.

Key idea

Critical values from the equation, then the sketch decides which pieces you keep.

Which method? Read the question

Read the question, then pick the tool.

You see	You do
$∣ a x + b ∣ = c$	Two cases: $a x + b = c$ or $a x + b = - c$
$∣ a x + b ∣ = c x + d$	Two cases, then check: reject any answer that makes $c x + d$ negative
$∣ a x + b ∣ = ∣ c x + d ∣$	Two cases, or square both sides
$∣ a x^{2} + b x + c ∣ = d$	Two quadratics. Up to four answers
$k ∣ a x + b ∣ < c$ or $> c$	Divide by k. Less than: sandwich. Greater than: two pieces
$k ∣ a x + b ∣ \leq ∣ c x + d ∣$	Square both sides, solve the quadratic inequality
$∣ a x + b ∣ \leq c x + d$ or a quadratic inside	Critical values, then test points or sketch

Common mistake

Giving the critical values as the final answer. An inequality question needs an inequality answer, such as $- 2 \leq x \leq 5$ .

Key idea

Every modulus question starts the same way: the modulus hides a sign, so split into two cases or square.

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