Solve for x
x\in (-\infty,-\frac{2}{3}]\cup [\frac{2}{3},\infty)
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x^{2}\geq \left(\frac{2}{3}\right)^{2}
Calculate the square root of \frac{4}{9} and get \frac{2}{3}. Rewrite \frac{4}{9} as \left(\frac{2}{3}\right)^{2}.
|x|\geq \frac{2}{3}
Inequality holds for |x|\geq \frac{2}{3}.
x\leq -\frac{2}{3}\text{; }x\geq \frac{2}{3}
Rewrite |x|\geq \frac{2}{3} as x\leq -\frac{2}{3}\text{; }x\geq \frac{2}{3}.
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