Modulus Equations and Inequalities

Welcome, students 👋 In this lesson, you will learn how to solve equations and inequalities involving the modulus, also called the absolute value. The modulus measures distance from zero on a number line, so it is always non-negative. That idea makes modulus problems very useful in real life, such as describing error tolerances, distances, and situations where a value can be above or below a target by the same amount. By the end of this lesson, you should be able to explain the meaning of modulus notation, solve equations such as $\lvert x \rvert = 5$, solve inequalities such as $\lvert x-3 \rvert < 4$, and connect these skills to graphs and function transformations.

What the modulus means

The modulus of a number $x$ is written as $\lvert x \rvert$. It gives the distance of $x$ from $0$ on the number line. Because distance cannot be negative, $\lvert x \rvert \ge 0$ for every real number $x$. For example, $\lvert 7 \rvert = 7$ and $\lvert -7 \rvert = 7$. This shows that two different numbers can have the same modulus if they are equally far from zero.

A useful way to understand modulus is through piecewise form:

$\lvert x \rvert =$

$\begin{cases}$

$ x, & x \ge 0 \\$

-x, & x < 0

$\end{cases}$

This definition is important because it lets you turn modulus equations and inequalities into simpler algebraic cases. Since IB Mathematics: Analysis and Approaches HL values clear reasoning, you should always show how you split the problem into cases when needed.

A real-world example is temperature difference. If a thermostat is set at $20^\circ\text{C}$ and the room temperature is $x$, then $\lvert x-20 \rvert$ tells you how far the temperature is from the target. If the acceptable difference is at most $2^\circ\text{C}$, then the condition becomes $\lvert x-20 \rvert \le 2$.

Solving modulus equations

A modulus equation usually asks you to find all values of a variable that make an absolute value statement true. The most common type is $\lvert f(x) \rvert = a$.

If $a < 0$, there is no solution, because a modulus cannot be negative. If $a \ge 0$, then

$\lvert f(x) \rvert = a$

means either

$f(x) = a$

$f(x) = -a.$

This is because a number and its opposite can both have the same distance from zero.

Example 1

Solve $\lvert x-4 \rvert = 7$.

Set up two equations:

$x-4 = 7$

$x-4 = -7.$

Solving gives

$x = 11$

$x = -3.$

So the solution set is $\{ -3, 11 \}$.

Example 2

Solve $\lvert 2x+1 \rvert = 3$.

Write the two cases:

$2x+1 = 3$

$2x+1 = -3.$

Then

2x = 2 \Rightarrow x = 1

and

2x = -4 \Rightarrow x = -2.

So the solutions are $x=1$ and $x=-2$.

A common mistake is forgetting the negative case. Another mistake is solving only one branch and missing the second valid answer. In IB exams, both branches matter unless the modulus equation is impossible because the right-hand side is negative.

Solving modulus inequalities

Modulus inequalities describe regions of values, not just individual solutions. They are especially important because they often represent intervals on a number line.

There are two main forms:

$\lvert f(x) \rvert < a$ or $\lvert f(x) \rvert \le a$
$\lvert f(x) \rvert > a$ or $\lvert f(x) \rvert \ge a$

When $a > 0$:

\lvert f(x) \rvert < a \quad \Longleftrightarrow \quad -a < f(x) < a

and

\lvert f(x) \rvert \le a \quad \Longleftrightarrow \quad -a \le f(x) \le a.

For the “greater than” cases:

\lvert f(x) \rvert > a \quad \Longleftrightarrow \quad f(x) < -a \text{ or } f(x) > a

and

\lvert f(x) \rvert \ge a \quad \Longleftrightarrow \quad f(x) \le -a \text{ or } f(x) \ge a.

These rules make sense because a value is within distance $a$ of zero only when it lies between $-a$ and $a$, and it is farther than $a$ from zero only when it lies outside that interval.

Example 3

Solve $\lvert x-3 \rvert < 4$.

Rewrite as a compound inequality:

-4 < x-3 < 4.

Add $3$ throughout:

-1 < x < 7.

So the solution is the interval $(-1,7)$.

Example 4

Solve $\lvert 3x-2 \rvert \ge 8$.

Split into two inequalities:

$3x-2 \le -8$

$3x-2 \ge 8.$

Solve each one:

3x \le -6 \Rightarrow x \le -2

and

3x \ge 10 \Rightarrow x \ge $\frac{10}{3}$.

So the solution set is $(-\infty,-2] \cup \left[\frac{10}{3},\infty\right)$.

Notice that the result is two separate intervals. This happens often with modulus inequalities involving “greater than” or “at least.”

Graphical understanding and function connections

Modulus problems are deeply connected to functions because $\lvert f(x) \rvert$ is itself a transformed version of the graph of $f(x)$. If $f(x)$ is negative, then $\lvert f(x) \rvert$ reflects that part of the graph above the $x$-axis. Positive parts stay unchanged.

For example, if $y=x-2$, then $y=\lvert x-2 \rvert$ becomes a V-shaped graph with vertex at $(2,0)$. This graph tells you a lot about equations and inequalities. Solving $\lvert x-2 \rvert = 3$ means finding where the graph meets the horizontal line $y=3$. The solutions are the $x$-values where the V-shape crosses that line.

This graph-based thinking is useful in IB Mathematics: Analysis and Approaches HL because it connects algebraic methods with function representation, one of the core ideas in the course. It also helps with interpreting solutions in context. If a measurement must stay within a range, then the graph shows the acceptable interval.

Modulus also appears in composite and inverse-type thinking. For example, if $g(x)=\lvert x \rvert$, then $g(f(x))=\lvert f(x) \rvert$. You may need to consider where $f(x)$ is positive or negative to understand the graph or solve related equations.

Strategy for solving difficult modulus problems

When a modulus problem looks complicated, use a systematic method:

Identify the expression inside the modulus.
Decide whether the problem is an equation or inequality.
For equations, split into two cases: $f(x)=a$ and $f(x)=-a$.
For inequalities, convert to a compound inequality or split into two outside regions.
Check your answers in the original statement.

Example 5

Solve $\lvert x^2-5x \rvert = 6$.

Set up two equations:

$x^2-5x=6$

$x^2-5x=-6.$

First equation:

$x^2-5x-6=0$

$(x-6)(x+1)=0$

so $x=6$ or $x=-1$.

Second equation:

$x^2-5x+6=0$

$(x-2)(x-3)=0$

so $x=2$ or $x=3$.

Therefore the full solution set is $\{ -1,2,3,6 \}$.

This example shows why algebraic structure matters. A modulus equation can lead to quadratic equations, so your polynomial skills are important here too.

Conclusion

Modulus equations and inequalities are a key part of Functions in IB Mathematics: Analysis and Approaches HL. The main idea is that $\lvert x \rvert$ represents distance, so it cannot be negative. Equations such as $\lvert f(x) \rvert = a$ are solved by considering both $f(x)=a$ and $f(x)=-a$, while inequalities such as $\lvert f(x) \rvert < a$ or $\lvert f(x) \rvert > a$ are converted into interval statements. These techniques connect algebra, graphs, and real-life interpretation. If you understand the meaning of modulus and practice careful case splitting, you will be ready to handle a wide range of IB questions confidently 🌟

Study Notes

$\lvert x \rvert$ means the distance from $x$ to $0$ on the number line.
$\lvert x \rvert \ge 0$ for all real $x$.
If $a < 0$, then $\lvert f(x) \rvert = a$ has no solution.
For $a \ge 0$, $\lvert f(x) \rvert = a$ means $f(x)=a$ or $f(x)=-a$.
$\lvert f(x) \rvert < a \Longleftrightarrow -a < f(x) < a$.
$\lvert f(x) \rvert \le a \Longleftrightarrow -a \le f(x) \le a$.
$\lvert f(x) \rvert > a \Longleftrightarrow f(x)<-a$ or $f(x)>a$.
$\lvert f(x) \rvert \ge a \Longleftrightarrow f(x)\le -a$ or $f(x)\ge a$.
Graphically, $y=\lvert f(x) \rvert$ keeps positive parts of $f(x)$ and reflects negative parts above the $x$-axis.
Always check answers in the original equation or inequality.