Solve for x
x\in (-\infty,-2\sqrt{10}]\cup [2\sqrt{10},\infty)
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x^{2}\geq \left(2\sqrt{10}\right)^{2}
Calculate the square root of 40 and get 2\sqrt{10}. Rewrite 40 as \left(2\sqrt{10}\right)^{2}.
|x|\geq 2\sqrt{10}
Inequality holds for |x|\geq 2\sqrt{10}.
x\leq -2\sqrt{10}\text{; }x\geq 2\sqrt{10}
Rewrite |x|\geq 2\sqrt{10} as x\leq -2\sqrt{10}\text{; }x\geq 2\sqrt{10}.
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