Example — two radicals compared

When both sides are radicals, the shortcut is immediate: one writes the existence conditions on both arguments and squares.

Example — $\sqrt{x^2-1}\ge \sqrt{x+5}$

Existence conditions: $\begin{cases}x^2-1\ge 0 \\ x+5\ge 0\end{cases} \iff \begin{cases}x\le -1 \vee x\ge 1 \\ x\ge -5\end{cases} \iff -5\le x\le -1 \vee x\ge 1.$

Squaring (legitimate because both sides are $\ge 0$): $x^2-1\ge x+5 \iff x^2-x-6\ge 0 \iff (x-3)(x+2)\ge 0 \iff x\le -2 \vee x\ge 3.$

Intersection with the existence conditions:

• with $-5\le x\le -1$: the region $x\le -2$, that is $-5\le x\le -2$;
• with $x\ge 1$: the region $x\ge 3$.

Solution: $\boxed{\,-5\le x\le -2 \ \vee\ x\ge 3\,}$.

The delicate step is not the squaring (free here), but remembering to intersect the result with the existence conditions: without them one would include values where one of the two roots does not exist.

Links

Topics: Irrational inequalities
Concepts: Existence conditions · Irrational inequality · Squaring
Methods: Non-negative sides shortcut
Skills: Solving inequalities