# Modulus Equations and Inequalities

> CIE IGCSE Additional Mathematics · 2025-2027
> Source: https://www.owlsprep.com/study/cie-0606-u4-modulus-equations-and-inequalities/

This guide covers solving linear modulus equations and inequalities for CIE IGCSE Additional Mathematics (0606). You will learn algebraic and graphical methods, plus how to identify and reject extraneous roots.

**Prerequisites:** Solving linear equations and inequalities; Sketching straight line graphs

## Learning objectives

- Solve modulus equations of the form |ax + b| = c, |ax + b| = g(x) and |ax² + bx + c| = d
- Identify and reject extraneous roots generated by squaring modulus equations
- Sketch modulus graphs with correct cusps and use them to solve equations/inequalities
- Solve modulus inequalities of both |ax + b| and |ax² + bx + c| types using algebraic and graphical methods

## Solving Modulus Equations Algebraically

**Modulus Equation** — An equation containing a modulus (absolute value) expression, for example $|ax + b| = k$ or $|ax^2 + bx + c| = d$, 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:* $|3x - 4| = 8$ 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.

## Sketching Modulus Graphs

Graphs of $y = |ax + b|$ 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.

**Cusp of a Modulus Graph** — The sharp turning point of $y=|ax + b|$, located at the coordinates $\left(-\frac{b}{a}, 0\right)$

*Example:* The graph of $y=|2x - 4|$ has a cusp at $(2, 0)$

**Worked example:** Sketch the graph of $y = |2x + 1|$, labelling the cusp and y-intercept

1. Find the cusp: set $2x + 1 = 0$ 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 $y=2x+1$; for $x < -0.5$, draw $y = -2x -1$, forming a sharp V-shape pointing upwards

> **tip**
>
> The solutions to $|ax + b| = g(x)$ are the x-coordinates of the intersection points between $y=|ax + b|$ and $y=g(x)$ on a graph.

## Solving Modulus Inequalities

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 $|3x + 2| \leq 7$

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: $y=|3x+2|$ is below $y=7$ between $x=-3$ and $x=\frac{5}{3}$, so the solution is correct.

> **tip**
>
> For a '>' modulus inequality $|f(x)| > g(x)$, any value of $x$ that makes the right-hand side negative is automatically a solution, because $|f(x)| \geq 0$ is always greater than a negative number. Only where $g(x) \geq 0$ do you need to split into cases and solve. (This shortcut does not apply to '=', '<' or '$\leq$' problems, where a negative right-hand side instead means no solution.)

**Worked example:** Solve $|x - 3| > 2x + 1$

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.

*Calculator:* allowed

## Identifying Extraneous Roots

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.

**Check your understanding**

1. You solve $|x + 1| = 2x - 3$ by squaring, and get solutions $x=4$ and $x=\frac{2}{3}$. Which solution is extraneous?

   *Why:* 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.

## The Modulus of a Quadratic Function

The modulus can also be applied to a quadratic. To sketch $y = |ax^2 + bx + c|$, first sketch the ordinary parabola $y = ax^2 + bx + c$, 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.

**Modulus of a Quadratic** — The graph of $y = |ax^2 + bx + c|$ is the parabola $y = ax^2 + bx + c$ 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 $x = -\sqrt{5}$ and $x = \sqrt{5}$ reflected upwards, giving a central peak of height 5 at $x = 0$.

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

> **tip**
>
> If a case such as $x^2 - 5 = -6$ leads to $x^2 = -1$, it has no real solutions, so simply discard that branch. This happens when $d$ is larger than the height of the reflected central peak.

> **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).

## Inequalities with a Modulus on Both Sides

A very common exam form compares two modulus expressions, such as $k|ax + b| \leq |cx + d|$ (for example $|5x + 4| \leq |2x - 3|$ 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.

**Two-Sided Modulus Inequality** — An inequality of the form $k|ax + b| \leq |cx + d|$ (or with $\geq$). Since $k|ax + b| \geq 0$ and $|cx + d| \geq 0$, squaring both sides is valid: $k^2(ax + b)^2 \leq (cx + d)^2$.

*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 $y = (3x + 2)(x - 6)$ 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 $|2x - 1| = |x + 4|$ 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): $(2x - 1)^2 = (x + 4)^2$.

   $$4x^2 - 4x + 1 = x^2 + 8x + 16 \\ 3x^2 - 12x - 15 = 0 \\ x^2 - 4x - 5 = 0 \\ (x - 5)(x + 1) = 0$$
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.

> **tip**
>
> The related equation $|ax + b| = |cx + d|$ is solved the same way: either square both sides, or write $ax + b = cx + d$ OR $ax + b = -(cx + d)$. Its solutions are exactly the x-coordinates where the two V-graphs cross, i.e. the endpoints of the inequality's solution interval. For example, $2|x - 1| = |x + 4|$ gives $x = -\frac{2}{3}$ and $x = 6$.

> **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.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Assuming $|ax + b| = c$ has solutions even when $c < 0$
  - Why it fails: The modulus function always outputs a non-negative value, so there are no real solutions if the RHS constant is negative.
  - Correct: First check if the RHS constant is non-negative; if not, state there are no real solutions.
- **Wrong:** Squaring both sides of a modulus equation and not checking solutions
  - Why it fails: Squaring removes the sign constraint of the modulus, so you can generate solutions that do not satisfy the original equation.
  - Correct: Substitute every solution from squaring back into the original equation, and reject any that do not match.
- **Wrong:** Drawing modulus graphs with a smooth curved minimum instead of a sharp cusp
  - Why it fails: 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: Mark the cusp point first, then draw two straight lines meeting at the cusp to form the V-shape.
- **Wrong:** Reversing inequality signs incorrectly for modulus inequalities
  - Why it fails: $|ax + b| < c$ corresponds to a bounded interval, while $|ax + b| > c$ corresponds to two separate unbounded intervals.
  - Correct: For $|f(x)| < c$ write $-c < f(x) < c$; for $|f(x)| > c$ write $f(x) < -c$ OR $f(x) > c$.
- **Wrong:** Forgetting to check if the RHS of a modulus equation is negative when it has x terms
  - Why it fails: The modulus output is always non-negative, so any solution where the RHS expression is negative is automatically invalid.
  - Correct: Either use the two-case method with sign checks, or use a graph to identify valid solution regions.

## Cheatsheet

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

## 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.

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/cie-0606-u4-modulus-equations-and-inequalities/