$$3x^{2}-9\ge0.$$
$x\in (-\infty,-\sqrt{3}]\cup [\sqrt{3},\infty)$
$$x^{2}\geq 3$$
$$x^{2}\geq \left(\sqrt{3}\right)^{2}$$
$$|x|\geq \sqrt{3}$$
$$x\leq -\sqrt{3}\text{; }x\geq \sqrt{3}$$
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