Modulus Equations and Inequalities

CIE IGCSE Additional Mathematics· 4.1, 4.2· 20 min read

1. Solving Modulus Equations Algebraically★★☆☆☆⏱ 5 min

📘 Definition

Modulus Equation

An equation containing a modulus (absolute value) expression, for example $∣ a x + b ∣ = k$ or $∣ a x^{2} + b x + c ∣ = d$ , where the right-hand side is a non-negative constant or a variable expression. If the right-hand side is a negative constant, there are no real solutions.

Example:

$∣3 x - 4∣ = 8$ or $∣ x^{2} - 5∣ = 4$

There are two valid algebraic methods to solve modulus equations: split into two cases based on the sign of the expression inside the modulus, or square both sides of the equation. If you use the squaring method, you must check all solutions against the original equation to eliminate extraneous roots.

📐 Worked Example

Solve |3x - 4| = 8

1
Method 1: Split into two cases
$3x - 4 = 8 \text{ (case 1: inner expression non-negative)}$
2
$3x = 12 \\ x = 4$
3
$-(3x - 4) = 8 \text{ (case 2: inner expression negative)}$
4
$-3x + 4 = 8 \\ -3x = 4 \\ x = -\frac{4}{3}$
5
Verify both solutions in the original equation: both are valid, so final solutions are $x = 4$ and $x = - \frac{4}{3}$

Often the modulus is not sitting by itself on one side. When you meet an equation such as $a ∣ x + b ∣ + c = d$ , first isolate the modulus (get $∣ x + b ∣$ on its own) and only then split into two cases. Never split while other terms are still attached to the modulus.

📐 Worked Example

Solve $2∣ x - 3∣ + 1 = 9$

1
The modulus is not alone, so isolate it first: subtract 1 from both sides, then divide by 2.
$2|x - 3| = 8 \\ |x - 3| = 4$
2
Now the modulus equals a non-negative constant, so split into two cases.
$x - 3 = 4 \quad \text{or} \quad x - 3 = -4$
3
$x = 7 \quad \text{or} \quad x = -1$
4
Check in the original equation: $2∣7 - 3∣ + 1 = 2 (4) + 1 = 9$ and $2∣ - 1 - 3∣ + 1 = 2 (4) + 1 = 9$ . Both are valid, so $x = 7$ or $x = - 1$ .

📐 Worked Example

Solve |x + 2| = 3x - 1 using the squaring method

1
Square both sides of the equation
$(x + 2)^2 = (3x - 1)^2$
2
$x^2 + 4x + 4 = 9x^2 - 6x + 1 \\ 8x^2 - 10x - 3 = 0$
3
Solve the quadratic equation
$x = \frac{10 \pm \sqrt{100 + 96}}{16} = \frac{10 \pm 14}{16}$
4
$x = 1.5 \text{ or } x = -0.25$
5
Check solutions in original equation: $x = 1.5$ is valid, $x = - 0.25$ gives LHS=1.75, RHS=-1.75 which is invalid, so reject $x = - 0.25$ . Final solution: $x = 1.5$

Exam tip:

Always label which method you are using to make your working clear for examiners.

2. Sketching Modulus Graphs★★★☆☆⏱ 5 min

Graphs of $y = ∣ a x + b ∣$ are V-shaped, with a sharp cusp at the point where the expression inside the modulus equals zero. All points on the graph lie on or above the x-axis, since the modulus output is always non-negative.

📘 Definition

Cusp of a Modulus Graph

The sharp turning point of $y = ∣ a x + b ∣$ , located at the coordinates $(- \frac{b}{a}, 0)$

Example:

The graph of $y = ∣2 x - 4∣$ has a cusp at $(2, 0)$

📐 Worked Example

Sketch the graph of $y = ∣2 x + 1∣$ , labelling the cusp and y-intercept

1
Find the cusp: set $2 x + 1 = 0$ so $x = - 0.5$ , cusp coordinates $(- 0.5, 0)$
2
Find the y-intercept: substitute $x = 0$ , $y = ∣1∣ = 1$ , so y-intercept at $(0, 1)$
3
Draw two straight lines: for $x \geq - 0.5$ , draw $y = 2 x + 1$ ; for $x < - 0.5$ , draw $y = - 2 x - 1$ , forming a sharp V-shape pointing upwards

3. Solving Modulus Inequalities★★★☆☆⏱ 6 min

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Modulus inequalities can be solved either with algebraic case analysis, or by sketching graphs and identifying the region that satisfies the inequality. Both methods are accepted in exams, as long as your working is clear.

📐 Worked Example

Solve $∣3 x + 2∣ \leq 7$

1
Rewrite the modulus inequality as a bounded double inequality
$-7 \leq 3x + 2 \leq 7$
2
$Subtract 2 from all parts: -9 \leq 3x \leq 5$
3
$Divide all parts by 3: -3 \leq x \leq \frac{5}{3}$
4
Verify with a graph: $y = ∣3 x + 2∣$ is below $y = 7$ between $x = - 3$ and $x = \frac{5}{3}$ , so the solution is correct.

📐 Worked Example

Solve $∣ x - 3∣ > 2 x + 1$

1
Case 1: $x \geq 3$ , so $∣ x - 3∣ = x - 3$
$x - 3 > 2x + 1 \\ x < -4$
2
No solutions in this range, since $x \geq 3$ and $x < - 4$ is impossible
3
Case 2: $x < 3$ , so $∣ x - 3∣ = - (x - 3) = - x + 3$
$-x + 3 > 2x + 1 \\ 2 > 3x \\ x < \frac{2}{3}$
4
This range is valid for $x < 3$ , so final solution is $x < \frac{2}{3}$

Exam tip:

For inequalities with a variable on the RHS, the graphical method is often more reliable to avoid missing invalid regions.

4. Identifying Extraneous Roots★★☆☆☆⏱ 4 min

Extraneous roots appear when you square both sides of an equation, because squaring converts negative values to positive, removing the sign constraint imposed by the modulus. You will lose marks if you do not reject these invalid solutions.

✓ Quick check

You solve $∣ x + 1∣ = 2 x - 3$ by squaring, and get solutions $x = 4$ and $x = \frac{2}{3}$ . Which solution is extraneous?
Reveal answer
$x=\frac{2}{3}$ —
Substitute $x = \frac{2}{3}$ into the original equation: LHS= $∣ \frac{5}{3} ∣ = \frac{5}{3}$ , RHS= $2 (\frac{2}{3}) - 3 = - \frac{5}{3}$ . The values are not equal, so $x = \frac{2}{3}$ is invalid.

5. The Modulus of a Quadratic Function★★★☆☆⏱ 6 min

The modulus can also be applied to a quadratic. To sketch $y = ∣ a x^{2} + b x + c ∣$ , first sketch the ordinary parabola $y = a x^{2} + b x + c$ , then reflect any part that lies below the x-axis up across the x-axis, so the whole graph is on or above the axis. To solve an equation or inequality, deal with each branch separately.

📘 Definition

Modulus of a Quadratic

The graph of $y = ∣ a x^{2} + b x + c ∣$ is the parabola $y = a x^{2} + b x + c$ with every part below the x-axis reflected upwards, so that all y-values are non-negative. The x-intercepts (the roots of the quadratic) are unchanged.

Example:

For $y = ∣ x^{2} - 5∣$ , the parabola $y = x^{2} - 5$ (minimum $(0, - 5)$ ) has the section between $x = - 5$ and $x = 5$ reflected upwards, giving a central peak of height 5 at $x = 0$ .

To solve $∣ a x^{2} + b x + c ∣ = d$ (with $d \geq 0$ ), split into the two quadratic equations $a x^{2} + b x + c = d$ and $a x^{2} + b x + c = - d$ , solve each, and keep every real root (one branch may have no real roots). For inequalities, rewrite $∣ a x^{2} + b x + c ∣ \leq d$ as the double inequality $- d \leq a x^{2} + b x + c \leq d$ , and rewrite $∣ a x^{2} + b x + c ∣ > d$ as $a x^{2} + b x + c > d$ OR $a x^{2} + b x + c < - d$ , then solve the resulting quadratic inequalities.

📐 Worked Example

Solve $∣ x^{2} - 5∣ = 4$

1
The right-hand side is $4 \geq 0$ , so split into two cases.
$x^2 - 5 = 4 \quad \text{or} \quad x^2 - 5 = -4$
2
First case:
$x^2 - 5 = 4 \\ x^2 = 9 \\ x = \pm 3$
3
Second case:
$x^2 - 5 = -4 \\ x^2 = 1 \\ x = \pm 1$
4
Both cases give real roots, so the four solutions are $x = - 3, - 1, 1, 3$ . Graphically these are where the line $y = 4$ meets $y = ∣ x^{2} - 5∣$ : twice on the outer arms and twice on the reflected central peak (which rises to height 5).

Exam tip:

Treat each of the two cases as its own quadratic, and always discard a case that gives no real roots (a negative number that would need a square root).

6. Inequalities with a Modulus on Both Sides★★★☆☆⏱ 6 min

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A very common exam form compares two modulus expressions, such as $k ∣ a x + b ∣ \leq ∣ c x + d ∣$ (for example $∣5 x + 4∣ \leq ∣2 x - 3∣$ or $2∣ x - 1∣ \leq ∣ x + 4∣$ ). Because every modulus is non-negative, both sides are non-negative, so you may square both sides without changing the direction of the inequality. This turns the problem into an ordinary quadratic inequality. You can also solve it graphically by drawing both V-shaped graphs on one set of axes and reading off where one lies below the other.

📘 Definition

Two-Sided Modulus Inequality

An inequality of the form $k ∣ a x + b ∣ \leq ∣ c x + d ∣$ (or with $\geq$ ). Since $k ∣ a x + b ∣ \geq 0$ and $∣ c x + d ∣ \geq 0$ , squaring both sides is valid: $k^{2} (a x + b)^{2} \leq (c x + d)^{2}$ .

Example:

$2∣ x - 1∣ \leq ∣ x + 4∣$ squares to $4 (x - 1)^{2} \leq (x + 4)^{2}$ .

📐 Worked Example

Solve $2∣ x - 1∣ \leq ∣ x + 4∣$ using (a) a graphical method and (b) an algebraic method.

1
(a) Graphical method: sketch both graphs on the same axes. $y = 2∣ x - 1∣$ is a narrow V with cusp $(1, 0)$ and slopes $\pm 2$ ; $y = ∣ x + 4∣$ is a wider V with cusp $(- 4, 0)$ and slopes $\pm 1$ .
2
Find where the graphs meet by equating the relevant straight-line pieces. Left intersection (here $x < 1$ so $2∣ x - 1∣ = 2 (1 - x)$ , and $x > - 4$ so $∣ x + 4∣ = x + 4$ ):
$2(1 - x) = x + 4 \\ 2 - 2x = x + 4 \\ -2 = 3x \\ x = -\frac{2}{3}$
3
Right intersection (here $x > 1$ so $2∣ x - 1∣ = 2 (x - 1)$ , and $x > - 4$ so $∣ x + 4∣ = x + 4$ ):
$2(x - 1) = x + 4 \\ 2x - 2 = x + 4 \\ x = 6$
4
From the sketch, the narrow V $y = 2∣ x - 1∣$ lies on or below the wide V $y = ∣ x + 4∣$ between the two intersection points, so the solution is $- \frac{2}{3} \leq x \leq 6$ .
5
(b) Algebraic method: both sides are non-negative, so square both sides.
$(2|x - 1|)^2 \leq |x + 4|^2 \\ 4(x - 1)^2 \leq (x + 4)^2$
6
$4(x^2 - 2x + 1) \leq x^2 + 8x + 16 \\ 4x^2 - 8x + 4 \leq x^2 + 8x + 16$
7
Bring all terms to one side to form a quadratic inequality, then factorise.
$3x^2 - 16x - 12 \leq 0 \\ (3x + 2)(x - 6) \leq 0$
8
The parabola $y = (3 x + 2) (x - 6)$ opens upwards and is $\leq 0$ between its roots $x = - \frac{2}{3}$ and $x = 6$ , so the solution is $- \frac{2}{3} \leq x \leq 6$ , matching the graphical method.
9
Check a value inside, $x = 0$ : LHS $= 2∣ - 1 ∣ = 2$ , RHS $= ∣4∣ = 4$ , and $2 \leq 4$ is true. Check a value outside, $x = 7$ : LHS $= 2∣6∣ = 12$ , RHS $= ∣11∣ = 11$ , and $12 \leq 11$ is false. The solution $- \frac{2}{3} \leq x \leq 6$ is confirmed.

📐 Worked Example

Solve $∣2 x - 1∣ = ∣ x + 4∣$ using (a) the two-case method and (b) the squaring method.

1
(a) Two-case method: when both sides are moduli, drop the bars and set the inside of the left equal to the right, then equal to the negative of the right.
$2x - 1 = x + 4 \quad \text{or} \quad 2x - 1 = -(x + 4)$
2
Solve the first equation.
$2x - 1 = x + 4 \\ x = 5$
3
Solve the second equation.
$2x - 1 = -x - 4 \\ 3x = -3 \\ x = -1$
4
(b) Squaring method (both sides are non-negative, so this is safe and needs no extraneous-root check): $(2 x - 1)^{2} = (x + 4)^{2}$ .
$4x^2 - 4x + 1 = x^2 + 8x + 16 \\ 3x^2 - 12x - 15 = 0 \\ x^2 - 4x - 5 = 0 \\ (x - 5)(x + 1) = 0$
5
Both methods give the same answers, $x = 5$ and $x = - 1$ . Check $x = 5$ : $∣2 (5) - 1∣ = 9 = ∣5 + 4∣$ . Check $x = - 1$ : $∣2 (- 1) - 1∣ = 3 = ∣ - 1 + 4∣$ . Both are valid.

Exam tip:

When both sides are a modulus (or a positive multiple of one), squaring is safe because both sides are non-negative, so no extraneous roots are introduced and you do not need the sign checks required when only one side is a modulus.

7. Common Pitfalls

Wrong move:

Assuming $∣ a x + b ∣ = c$ has solutions even when $c < 0$

Why:

The modulus function always outputs a non-negative value, so there are no real solutions if the RHS constant is negative.

Correct move:

First check if the RHS constant is non-negative; if not, state there are no real solutions.

Wrong move:

Squaring both sides of a modulus equation and not checking solutions

Why:

Squaring removes the sign constraint of the modulus, so you can generate solutions that do not satisfy the original equation.

Correct move:

Substitute every solution from squaring back into the original equation, and reject any that do not match.

Wrong move:

Drawing modulus graphs with a smooth curved minimum instead of a sharp cusp

Why:

The modulus graph changes direction abruptly at the point where the inner expression is zero, so it has a sharp corner, not a smooth curve.

Correct move:

Mark the cusp point first, then draw two straight lines meeting at the cusp to form the V-shape.

Wrong move:

Reversing inequality signs incorrectly for modulus inequalities

Why:

$∣ a x + b ∣ < c$ corresponds to a bounded interval, while $∣ a x + b ∣ > c$ corresponds to two separate unbounded intervals.

Correct move:

For $∣ f (x) ∣ < c$ write $- c < f (x) < c$ ; for $∣ f (x) ∣ > c$ write $f (x) < - c$ OR $f (x) > c$ .

Wrong move:

Forgetting to check if the RHS of a modulus equation is negative when it has x terms

Why:

The modulus output is always non-negative, so any solution where the RHS expression is negative is automatically invalid.

Correct move:

Either use the two-case method with sign checks, or use a graph to identify valid solution regions.

8. Quick Reference Cheatsheet

Problem Type	Method	Key Check
$∣ a x + b ∣ = c$ ( $c \geq 0$ constant)	Split into $a x + b = c$ and $a x + b = - c$ , solve	No check needed for constant $c \geq 0$
$∣ a x + b ∣ = g (x)$ (variable RHS)	Square both sides, solve quadratic	Check all solutions in original equation, reject extraneous roots
$∣ a x + b ∣ < c$ ( $c > 0$ constant)	Rewrite as $- c < a x + b < c$ , solve	Preserve inequality direction when dividing/multiplying
$∣ a x + b ∣ > c$ ( $c > 0$ constant)	Rewrite as $a x + b < - c$ OR $a x + b > c$ , solve	Do not combine the two separate solution intervals
Sketch $y = ∣ a x + b ∣$	Mark cusp at $x = - \frac{b}{a}$ , draw two straight lines	Ensure sharp cusp, no smooth curved minimum

9. Frequently Asked

Do I always need to check solutions when squaring modulus equations?

Yes! Squaring removes the sign constraint of the modulus, so you must substitute every solution back into the original equation to eliminate extraneous roots, otherwise you will lose marks.

Can I use graphical methods to solve modulus inequalities even if not specified?

Absolutely, as long as your graph is accurately drawn with labelled axes and correct cusps, it is a fully valid method for both equations and inequalities in exams.

Going deeper

syllabusCIE 0606 2025-2027 Official SyllabusRefer to section 4.1 and 4.2 for official topic guidance

What's Next

Now that you have mastered modulus equations and inequalities, you can apply these skills to more complex topics in CIE IGCSE Additional Mathematics 0606, including solving systems of equations involving modulus, and analyzing intersections between modulus graphs and quadratic or cubic functions. You will also need this knowledge for coordinate geometry problems involving distance calculations, and for calculus topics with absolute value functions. This topic is frequently tested in both Paper 1 (non-calculator) and Paper 2 (calculator), so practice both algebraic and graphical methods to maximize your marks.