## College Algebra (10th Edition)

$x\leq-3$ or $x\geq 3$
We know that the solution to the inequality: $x^2\gt a$ Is equivalent to: $x\lt -\sqrt{a}$ or $x\gt \sqrt{a}$ Therefore: $x^{2}\geq 9$ $x\leq-\sqrt{9}$ or $x\geq\sqrt{9}$ $x\leq-3$ or $x\geq 3$