Solve for x
x\in (-\infty,-\frac{4}{3}]\cup [\frac{4}{3},\infty)
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x^{2}\geq \frac{16}{9}
Add \frac{16}{9} to both sides.
x^{2}\geq \left(\frac{4}{3}\right)^{2}
Calculate the square root of \frac{16}{9} and get \frac{4}{3}. Rewrite \frac{16}{9} as \left(\frac{4}{3}\right)^{2}.
|x|\geq \frac{4}{3}
Inequality holds for |x|\geq \frac{4}{3}.
x\leq -\frac{4}{3}\text{; }x\geq \frac{4}{3}
Rewrite |x|\geq \frac{4}{3} as x\leq -\frac{4}{3}\text{; }x\geq \frac{4}{3}.
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