Special case - modulus less than or equal to B(x)

When the inequality has the form $|A(x)|\le B(x)$ (a modulus less than something), there is a shortcut that avoids the splitting: you turn it into a system with a double inequality.

Property — $|A(x)|\le B(x)$

The inequality $|A(x)|\le B(x)$ (or with $<$) is equivalent to the system $\begin{cases} B(x)\ge0 \\ -B(x)\le A(x)\le B(x) \end{cases}$ where the first condition $B(x)\ge0$ is necessary: an absolute value, which is always $\ge0$, cannot be less than a negative number. If $B(x)<0$ the inequality is impossible.

Observation — Why the double inequality

Under the condition $B(x)\ge0$, writing $|A(x)|\le B(x)$ means ”$A(x)$ is no further than $B(x)$ from zero”, that is $-B(x)\le A(x)\le B(x)$. It is exactly like solving $|x-3|\le5$: it is equivalent to $-5\le x-3\le5$, whence $-2\le x\le8$.

Example

$|2x-1|\le3$. Here $A(x)=2x-1$ and $B(x)=3$ is a positive constant, so the condition $B\ge0$ is automatic. From the double inequality: $-3\le2x-1\le3 \iff -2\le2x\le4 \iff \boxed{-1\le x\le2}.$

Example

$|x^2-4|\le x+2$. Here $A(x)=x^2-4$ and $B(x)=x+2$. System: $\begin{cases} x+2\ge0 \\ -(x+2)\le x^2-4\le x+2 \end{cases}$

• First condition: $x\ge-2$.
• Right-hand part: $x^2-4\le x+2 \iff x^2-x-6\le0 \iff -2\le x\le3$.
• Left-hand part: $-(x+2)\le x^2-4 \iff x^2+x-2\ge0 \iff x\le-2\ \lor\ x\ge1$.

Intersecting the three conditions gives $\{x=-2\}\cup[1,3]$.

Solution: $\boxed{x=-2\ \lor\ 1\le x\le3}$.

Links

Topics: Parabola
Concepts: Inequalities with absolute value · Absolute value
Skills: Solving inequalities · Solving systems