Inequalities involving absolute value

The absolute value $|x|$ measures the distance of $x$ from $0$ on the number line. Inequalities involving $|x|$ are inequalities about distance , and they almost always reduce to one or two linear inequalities.

The two master rules

For any $a > 0$:

$\boxed{|x| < a \iff -a < x < a.}$

$\boxed{|x| > a \iff x < -a \text{ or } x > a.}$

The first describes "within distance $a$ from $0$" , a bounded interval $(-a, a)$. The second describes "more than distance $a$ from $0$" , the complement, $(-\infty, -a) \cup (a, \infty)$.

For $a = 0$: $|x| < 0$ has no solution; $|x| > 0$ means $x \ne 0$; $|x| \le 0$ means $x = 0$.

With non-strict inequalities, the brackets close: $|x| \le a \iff -a \le x \le a$, and $|x| \ge a \iff x \le -a$ or $x \ge a$.

Inequalities of the form $|x - c| < a$ or $|x - c| > a$

Shift by $c$. $|x - c|$ is the distance from $x$ to $c$. So:

$|x - c| < a \iff -a < x - c < a \iff c - a < x < c + a.$

This is the interval centred at $c$ with radius $a$. Similarly $|x - c| \ge a$ describes the points at distance $\ge a$ from $c$.

Inequalities with $|f(x)| < g(x)$

If $g(x) > 0$, then $|f(x)| < g(x) \iff -g(x) < f(x) < g(x)$. Solve the two linear (or polynomial) inequalities and intersect.

If $g(x)$ is itself an unknown, you must also ensure $g(x) > 0$.

Triangle inequality

$|x + y| \le |x| + |y|$ for all real $x, y$. With equality iff $x$ and $y$ have the same sign (or one is zero). The reverse: $\big||x| - |y|\big| \le |x - y|$.

These hold for real numbers, but they also generalise to complex numbers (Chapter 4) and to vectors (Class XII).

Worked examples

Example 1. Solve $|x - 2| < 3$.

$-3 < x - 2 < 3 \iff -1 < x < 5$. Solution: $(-1, 5)$.

Example 2. Solve $|2x + 1| \ge 5$.

$2x + 1 \ge 5$ or $2x + 1 \le -5$. First: $x \ge 2$. Second: $x \le -3$. Solution: $(-\infty, -3] \cup [2, \infty)$.

Example 3. Solve $|x - 3| < |x + 1|$.

Square both sides (both non-negative, so direction preserved): $(x - 3)^2 < (x + 1)^2 \iff x^2 - 6x + 9 < x^2 + 2x + 1 \iff -6x + 9 < 2x + 1 \iff 8 < 8x \iff x > 1.$ Solution: $(1, \infty)$.

(Geometric check: $|x - 3|$ is distance to $3$, $|x + 1|$ is distance to $-1$. The midpoint of $3$ and $-1$ is $1$; points closer to $3$ have $x > 1$.)

Example 4. Solve $|x| + |x - 2| < 3$.

Case-split by sign of $x$ and $x - 2$:

• $x < 0$: $|x| = -x$, $|x - 2| = 2 - x$. Sum: $-x + 2 - x = 2 - 2x < 3 \iff x > -\tfrac{1}{2}$. Intersect with $x < 0$: $(-\tfrac{1}{2}, 0)$.
• $0 \le x \le 2$: $|x| = x$, $|x - 2| = 2 - x$. Sum: $x + 2 - x = 2 < 3$, always true. Solution: $[0, 2]$.
• $x > 2$: $|x| = x$, $|x - 2| = x - 2$. Sum: $2x - 2 < 3 \iff x < \tfrac{5}{2}$. Intersect: $(2, \tfrac{5}{2})$.

Union: $(-\tfrac{1}{2}, \tfrac{5}{2})$.

Example 5 (harder). Solve $\dfrac{|x - 1|}{x + 2} \le 1$.

Case 1: $x + 2 > 0$, i.e. $x > -2$. Inequality $\iff |x - 1| \le x + 2$.

If $x \ge 1$: $|x - 1| = x - 1 \le x + 2$, always true.

If $x < 1$ (and $x > -2$): $|x - 1| = 1 - x \le x + 2 \iff -1 \le 2x \iff x \ge -\tfrac{1}{2}$.

So Case 1: $x \ge -\tfrac{1}{2}$.

Case 2: $x + 2 < 0$, i.e. $x < -2$. Then $|x - 1| \ge 0 > x + 2$ is negative, so $\dfrac{|x - 1|}{x + 2}$ is $\le 0 < 1$, unless the modulus is zero (it cannot be in this range since $x = 1 > -2$). So the inequality is automatically satisfied: $x < -2$.

Combining: $(-\infty, -2) \cup [-\tfrac{1}{2}, \infty)$.

Try it yourself

1. Solve $|x| \le 4$.
2. Solve $|x + 3| > 2$.
3. Solve $|2x - 5| \le 7$.
4. Solve $|x - 1| < |x + 2|$.
5. Solve $|x| + |x - 1| \le 3$.
6. Solve $\dfrac{|x| - 1}{|x| + 1} < \dfrac{1}{2}$.
7. Solve $\big| x - \tfrac{1}{2}\big| < \tfrac{1}{4}$.
8. Solve $|3x + 5| \ge 4$.
9. Solve $|x^2 - 1| < 3$ (reduces to a quadratic-in-$x$ inequality).
10. Solve $\big|x - 2| - |x + 1\big| \ge 1$.
11. Show that the solution set of $|x - 1| + |x + 1| \le 2$ is exactly $[-1, 1]$.
12. Solve $|x - 1|^2 - 2|x - 1| - 3 \ge 0$ (a quadratic in $|x - 1|$).

Pitfalls / Tricks

• $|x| < a$ produces a single interval; $|x| > a$ produces two intervals. Do not confuse.
• For $|f(x)| < g(x)$, always check $g(x) > 0$ first , otherwise no solutions are possible.
• Squaring both sides of $|f(x)| < |g(x)|$ is safe because both sides are $\ge 0$.
• Insight. Absolute value is just "distance". Every absolute-value inequality is a distance statement, and is often easiest to read off a number line.