Solve for x
x\in (-\infty,-\frac{\sqrt{6}}{3}]\cup [\frac{\sqrt{6}}{3},\infty)
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x^{2}\geq \frac{2}{3}
Add \frac{2}{3} to both sides.
x^{2}\geq \left(\frac{\sqrt{6}}{3}\right)^{2}
Calculate the square root of \frac{2}{3} and get \frac{\sqrt{6}}{3}. Rewrite \frac{2}{3} as \left(\frac{\sqrt{6}}{3}\right)^{2}.
|x|\geq \frac{\sqrt{6}}{3}
Inequality holds for |x|\geq \frac{\sqrt{6}}{3}.
x\leq -\frac{\sqrt{6}}{3}\text{; }x\geq \frac{\sqrt{6}}{3}
Rewrite |x|\geq \frac{\sqrt{6}}{3} as x\leq -\frac{\sqrt{6}}{3}\text{; }x\geq \frac{\sqrt{6}}{3}.
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