Modulus Equations and Inequalities

The modulus $\lvert x \rvert$ measures distance from zero, so it turns up whenever size matters but direction doesn’t: tolerances in manufacturing, errors in measurement, “within 5 of”. This page shows you how to solve equations and inequalities that contain a modulus, by splitting into cases, by squaring, and by reading a graph. It finishes with the general skill of solving $g(x) \ge f(x)$, both on a graph and with algebra.

Key ideas

The modulus as a distance

$$\lvert x \rvert = \begin{cases} x, & x \ge 0 \\ -x, & x \lt 0 \end{cases}$$

$\lvert x \rvert$ is the distance from $x$ to $0$, and $\lvert x - a \rvert$ is the distance from $x$ to $a$. Some useful facts:

$$\lvert x \rvert \ge 0, \qquad \lvert ab \rvert = \lvert a \rvert \lvert b \rvert, \qquad \lvert x \rvert^2 = x^2, \qquad \sqrt{x^2} = \lvert x \rvert$$

Basic equations and inequalities

For a number $k \gt 0$:

Statement	Means	Solution
$\lvert x \rvert = k$	distance from $0$ is $k$	$x = k$ or $x = -k$
$\lvert x \rvert \lt k$	closer than $k$ to $0$	$-k \lt x \lt k$
$\lvert x \rvert \gt k$	further than $k$ from $0$	$x \lt -k$ or $x \gt k$

The same works with any expression inside: $\lvert f(x) \rvert = k$ means $f(x) = k$ or $f(x) = -k$. If $k \lt 0$, the equation $\lvert f(x) \rvert = k$ has no solutions, because a modulus can’t be negative.

Method 1: cases

Split according to the sign of what’s inside the modulus.

• $\lvert f(x) \rvert = \lvert g(x) \rvert$: solve $f(x) = g(x)$ and $f(x) = -g(x)$. Both sides are moduli, so every solution works.
• $\lvert f(x) \rvert = g(x)$: solve $f(x) = g(x)$ and $f(x) = -g(x)$, then check each answer in the original equation. The right side $g(x)$ must be $\ge 0$, and any solution that makes it negative is extraneous.

Method 2: squaring

If both sides are non-negative, squaring keeps an equation or inequality true and removes the modulus signs:

$$\lvert f(x) \rvert \lt \lvert g(x) \rvert \quad\Longleftrightarrow\quad [f(x)]^2 \lt [g(x)]^2$$

This turns the problem into a polynomial one. It’s safe when both sides are moduli (or are otherwise known to be non-negative). If one side could be negative, use cases instead.

Method 3: graphs

To solve $\lvert f(x) \rvert = g(x)$ or $\lvert f(x) \rvert \gt g(x)$, sketch $y = \lvert f(x) \rvert$ and $y = g(x)$ (see modulus graphs). The intersection points give the equation’s solutions, and the inequality holds wherever the first graph is above the second. A graph also tells you how many solutions to expect, which is a great check on the algebra.

Solving g(x) ≥ f(x)

Any inequality between two functions can be handled the same way.

• Graphically: draw $y = g(x)$ and $y = f(x)$, find their intersection points (by hand or with the GDC’s “intersect” tool), and read off the $x$-values where the graph of $g$ is on or above the graph of $f$.
• Analytically: rearrange to $g(x) - f(x) \ge 0$, factor, and use a sign chart, as on polynomial inequalities. For fractions, see rational inequalities.

The IB guide expects the algebraic method for simple polynomials up to degree $3$, and technology for other functions.

On the SAT

Desmos understands absolute value: type abs(2x - 1) (or use the | key). On the SAT, graph y = abs(2x - 1) and y = 5 and click the intersections: $x = -2$ and $x = 3$. For an inequality, read where the V-shaped graph is above or below the line. For simple equations like $|x - 4| = 6$, the two cases in your head ($x - 4 = 6$ or $x - 4 = -6$) are faster. See using Desmos on the SAT.

Worked examples

Example 1: A basic equation and inequality

Solve

• (a) $\lvert 2x - 5 \rvert = 7$
• (b) $\lvert 2x - 5 \rvert \lt 7$

Solution.

(a) $2x - 5 = 7$ or $2x - 5 = -7$. So $2x = 12$ or $2x = -2$, giving $x = 6$ or $x = -1$.

Check: $\lvert 12 - 5 \rvert = 7$ ✓ and $\lvert -2 - 5 \rvert = 7$ ✓.

(b) $\lvert 2x - 5 \rvert \lt 7$ means $-7 \lt 2x - 5 \lt 7$. Add $5$ to all three parts, then divide by $2$:

$$-2 \lt 2x \lt 12 \quad\Rightarrow\quad -1 \lt x \lt 6$$

Notice how (a) gives the endpoints of (b). Test a value inside: $x = 0$ gives $\lvert -5 \rvert = 5 \lt 7$ ✓.

Example 2: Rejecting an extraneous solution

Solve $\lvert x + 1 \rvert = 2x - 4$.

Solution. The right side isn’t a modulus, so use cases and check.

Case 1: $x + 1 = 2x - 4$, so $x = 5$. Check: $\lvert 6 \rvert = 6$ and $2(5) - 4 = 6$ ✓.

Case 2: $x + 1 = -(2x - 4)$, so $x + 1 = -2x + 4$, $3x = 3$, $x = 1$. Check: $\lvert 2 \rvert = 2$ but $2(1) - 4 = -2$. ✗ A modulus can’t equal $-2$, so reject $x = 1$.

The only solution is $x = 5$.

Why the extra answer? On a graph, $y = 2x - 4$ meets the V shape $y = \lvert x + 1 \rvert$ only once, on its right arm. The second case found where the line meets the extension of the left arm, $y = -(x + 1)$, below the $x$-axis, which isn’t part of the modulus graph.

Example 3: Squaring both sides

Solve $\lvert x - 3 \rvert \ge \lvert 2x + 1 \rvert$.

Solution. Both sides are non-negative, so square:

$$\begin{aligned} (x - 3)^2 &\ge (2x + 1)^2 \\ x^2 - 6x + 9 &\ge 4x^2 + 4x + 1 \\ 0 &\ge 3x^2 + 10x - 8 \\ 0 &\ge (3x - 2)(x + 4) \end{aligned}$$

The parabola $y = (3x - 2)(x + 4)$ opens up with zeros $-4$ and $\frac{2}{3}$, so it’s $\le 0$ between them:

$$-4 \le x \le \frac{2}{3}$$

$\lvert x - 3 \rvert \ge \lvert 2x + 1 \rvert$ where the blue graph is on or above the orange graph: from $x = -4$ to $x = \frac{2}{3}$.

Check the endpoints: $\lvert -7 \rvert = 7 = \lvert -7 \rvert$ ✓ and $\left\lvert -\frac{7}{3} \right\rvert = \frac{7}{3} = \left\lvert \frac{7}{3} \right\rvert$ ✓. Check a point inside: $x = 0$ gives $3 \ge 1$ ✓.

Example 4: Solving g(x) ≥ f(x) for a cubic

Let $f(x) = x^3 - 2x^2$ and $g(x) = 5x - 6$. Solve $g(x) \ge f(x)$ analytically, and confirm with a graph.

Solution. Rearrange so that one side is $0$:

$$5x - 6 \ge x^3 - 2x^2 \quad\Longleftrightarrow\quad x^3 - 2x^2 - 5x + 6 \le 0$$

Let $p(x) = x^3 - 2x^2 - 5x + 6$. Try $x = 1$: $p(1) = 1 - 2 - 5 + 6 = 0$, so $(x - 1)$ is a factor (factor theorem). Dividing gives

$$p(x) = (x - 1)(x^2 - x - 6) = (x - 1)(x + 2)(x - 3)$$

Sign chart with critical values $-2$, $1$ and $3$:

Interval	$x + 2$	$x - 1$	$x - 3$	$p(x)$
$x \lt -2$	$-$	$-$	$-$	$-$
$-2 \lt x \lt 1$	$+$	$-$	$-$	$+$
$1 \lt x \lt 3$	$+$	$+$	$-$	$-$
$x \gt 3$	$+$	$+$	$+$	$+$

We need $p(x) \le 0$, including the zeros:

$$x \le -2 \quad\text{or}\quad 1 \le x \le 3$$

Graphically: the line and cubic meet at $(-2, -16)$, $(1, -1)$ and $(3, 9)$, and the line is above the cubic exactly on those intervals.

$g(x) \ge f(x)$ where the orange line is on or above the blue cubic: $x \le -2$ or $1 \le x \le 3$.

Common mistakes

Splitting $\lvert x \rvert \lt k$ into two pieces. $\lvert x \rvert \lt k$ is one interval, $-k \lt x \lt k$. It’s $\lvert x \rvert \gt k$ that splits into two pieces.

Not checking solutions of $\lvert f(x) \rvert = g(x)$. When the right side isn’t a modulus, one of the cases can give an answer that makes $g(x)$ negative. Substitute every answer back into the original equation.

Squaring when one side might be negative. $\lvert x + 1 \rvert = 2x - 4$ squared gives $(x + 1)^2 = (2x - 4)^2$, which has solutions $x = 5$ and $x = 1$, and $x = 1$ is wrong. Squaring is only safe when both sides are known to be non-negative.

Dividing an inequality by an expression that could be negative. In $x^3 \ge 4x$, dividing by $x$ loses solutions and can flip the sign wrongly. Move everything to one side and factor instead.

Reading the wrong side of the graph. For $g(x) \ge f(x)$, you want the $x$-values where the graph of $g$ is above the graph of $f$. Write the answer as $x$-intervals, not $y$-values, and include the intersection points when the inequality is $\ge$ or $\le$.

Missing intersection points on the GDC. Two curves can meet at points very close together. If the graph looks like it nearly touches somewhere, zoom in before you decide there’s no intersection.

Practice

1. (Warm-up) Solve $\lvert x - 4 \rvert = 9$.

Solution

$x - 4 = 9$ or $x - 4 = -9$, so $x = 13$ or $x = -5$.

Check: $\lvert 9 \rvert = 9$ ✓ and $\lvert -9 \rvert = 9$ ✓.

2. (Warm-up) Solve

• (a) $\lvert 3x + 2 \rvert \le 8$
• (b) $\lvert 2x - 1 \rvert = -3$

Solution

(a) $-8 \le 3x + 2 \le 8$, so $-10 \le 3x \le 6$ and

$$-\frac{10}{3} \le x \le 2$$

(b) No solutions: a modulus is never negative.

3. (Core) Solve $\lvert x - 1 \rvert = \lvert 3x + 5 \rvert$.

Solution

Both sides are moduli, so either the insides are equal or they are opposites.

$x - 1 = 3x + 5$ gives $-2x = 6$, so $x = -3$.

$x - 1 = -(3x + 5)$ gives $4x = -4$, so $x = -1$.

Check: $\lvert -4 \rvert = \lvert -4 \rvert$ ✓ and $\lvert -2 \rvert = \lvert 2 \rvert$ ✓.

4. (Core) Solve $\lvert x^2 - 5 \rvert = 4$.

Solution

$x^2 - 5 = 4$ or $x^2 - 5 = -4$. So $x^2 = 9$ or $x^2 = 1$:

$$x = -3, \ -1, \ 1, \ 3$$

On a graph, the horizontal line $y = 4$ crosses the W shape $y = \lvert x^2 - 5 \rvert$ four times, which confirms there are four solutions.

5. (Core) Solve $\lvert 2x + 3 \rvert \gt x + 6$.

Solution

Split at $x = -\frac{3}{2}$, where $2x + 3$ changes sign.

Case $x \ge -\frac{3}{2}$: $2x + 3 \gt x + 6$ gives $x \gt 3$. All of these satisfy $x \ge -\frac{3}{2}$, so this case gives $x \gt 3$.

Case $x \lt -\frac{3}{2}$: $-(2x + 3) \gt x + 6$ gives $-3x \gt 9$, so $x \lt -3$ (dividing by $-3$ flips the sign). All of these satisfy $x \lt -\frac{3}{2}$.

$$x \lt -3 \quad\text{or}\quad x \gt 3$$

Check: $x = 0$ gives $3 \gt 6$, false ✓ (it’s not in the solution); $x = 4$ gives $11 \gt 10$ ✓; $x = -4$ gives $5 \gt 2$ ✓.

6. (Core) Let $f(x) = x^3 - x$ and $g(x) = 3x^2 - 3$. Solve $g(x) \ge f(x)$ analytically.

Solution

$$3x^2 - 3 \ge x^3 - x \quad\Longleftrightarrow\quad x^3 - 3x^2 - x + 3 \le 0$$

Factor by grouping: $x^2(x - 3) - (x - 3) = (x - 3)(x^2 - 1) = (x - 3)(x - 1)(x + 1)$.

Critical values $-1$, $1$, $3$. The cubic has a positive leading coefficient, so its signs from left to right are $-$, $+$, $-$, $+$. We need $\le 0$:

$$x \le -1 \quad\text{or}\quad 1 \le x \le 3$$

Check: $x = 2$ gives $g(2) = 9$ and $f(2) = 6$, and $9 \ge 6$ ✓. $x = 0$ gives $g(0) = -3$ and $f(0) = 0$, and $-3 \ge 0$ is false ✓.

7. (Core) A machine fills bags of flour with a target mass of $500$ g. A bag passes inspection if its mass $m$ grams satisfies $\lvert m - 500 \rvert \le 12$.

• (a) Find the range of masses that pass.
• (b) Does a bag of $487.5$ g pass?

Solution

(a) $-12 \le m - 500 \le 12$, so $488 \le m \le 512$ grams. In words: the mass must be within $12$ g of $500$ g.

(b) $\lvert 487.5 - 500 \rvert = 12.5 \gt 12$, so no, it fails.

8. (Challenge) Solve $\lvert x - 2 \rvert + \lvert x + 1 \rvert = 5$.

Solution

The insides change sign at $x = 2$ and $x = -1$, so use three cases.

$x \ge 2$: both insides are non-negative: $(x - 2) + (x + 1) = 5$, so $2x - 1 = 5$ and $x = 3$. This satisfies $x \ge 2$ ✓.

$-1 \le x \lt 2$: $-(x - 2) + (x + 1) = 5$ gives $3 = 5$, which is impossible. No solutions here.

$x \lt -1$: $-(x - 2) - (x + 1) = 5$ gives $-2x + 1 = 5$, so $x = -2$. This satisfies $x \lt -1$ ✓.

Solutions: $x = -2$ and $x = 3$.

This makes sense as distances: $\lvert x - 2 \rvert + \lvert x + 1 \rvert$ is the distance from $x$ to $2$ plus the distance from $x$ to $-1$. Any point between $-1$ and $2$ gives a total of exactly $3$, so the total is $5$ only at points $1$ unit beyond either end.

9. (Challenge) Use technology to solve $e^x \le \lvert x^2 - 4 \rvert$. Give the endpoints to 3 s.f.

Solution

Graph $y = e^x$ and $y = \lvert x^2 - 4 \rvert$ (on most GDCs, “abs”). The modulus graph is a W shape touching the axis at $x = \pm 2$, with a bump of height $4$ at $x = 0$.

Use “intersect” for every crossing. Near $x = -2$, zoom in: the two graphs cross twice very close together, because $e^{-2} \approx 0.135$ is small and the W shape dips to $0$ there.

$$x = -2.03, \qquad x = -1.96, \qquad x = 1.06$$

For $x \gt 2$, $e^x$ is always above $x^2 - 4$ (at $x = 2$ it’s $7.39$ against $0$, and it grows faster), so there are no more crossings.

The modulus graph is on or above the exponential:

$$x \le -2.03 \quad\text{or}\quad -1.96 \le x \le 1.06$$

Check a point in each part: $x = -3$ gives $e^{-3} \approx 0.050 \le 5$ ✓; $x = -2$ gives $0.135 \le 0$, false ✓ (excluded); $x = 0$ gives $1 \le 4$ ✓; $x = 2$ gives $7.39 \le 0$, false ✓.