Non-strict inequality

Statement

Solve $\sqrt{x^2-4}\ge x-3$.

Solution

The root is on the greater side (Type I), with a non-strict sign: a union of two cases.

Case (a): $\begin{cases}x^2-4\ge 0 \\ x-3 < 0\end{cases} \iff (x\le -2 \vee x\ge 2)\ \wedge\ x<3 \iff x\le -2 \ \vee\ 2\le x<3$.

Case (b): $\begin{cases}x-3\ge 0 \\ x^2-4\ge (x-3)^2\end{cases}$. The second: $x^2-4\ge x^2-6x+9 \iff 6x\ge 13 \iff x\ge \tfrac{13}{6}$. Intersecting with $x\ge 3$: $x\ge 3$.

Union: $\boxed{x\le -2 \ \vee\ x\ge 2}$.

Links

Topics: Radicals
Concepts: Irrational inequality
Methods: Irrational inequality case analysis
Skills: Reasoning by cases · Solving inequalities
Exercise type: Solving an inequality