Solve 32x^2geq1

$$32 x ^ { 2 } \geq 1$$

Solve for x

$$x\in (-\infty,-\frac{\sqrt{2}}{8}]\cup [\frac{\sqrt{2}}{8},\infty)$$

Solution Steps

Divide both sides by 32. Since 32 is positive, the inequality direction remains the same.

$$x^{2}\geq \frac{1}{32}$$

Calculate the square root of \frac{1}{32} and get \frac{\sqrt{2}}{8}. Rewrite \frac{1}{32} as \left(\frac{\sqrt{2}}{8}\right)^{2}.

$$x^{2}\geq \left(\frac{\sqrt{2}}{8}\right)^{2}$$

Inequality holds for |x|\geq \frac{\sqrt{2}}{8}.

$$|x|\geq \frac{\sqrt{2}}{8}$$

Rewrite |x|\geq \frac{\sqrt{2}}{8} as x\leq -\frac{\sqrt{2}}{8}\text{; }x\geq \frac{\sqrt{2}}{8}.

$$x\leq -\frac{\sqrt{2}}{8}\text{; }x\geq \frac{\sqrt{2}}{8}$$