Solve 81x^2geq16

$$81 x ^ { 2 } \geq 16$$

Solve for x

$$x\in (-\infty,-\frac{4}{9}]\cup [\frac{4}{9},\infty)$$

Solution Steps

Divide both sides by 81. Since 81 is positive, the inequality direction remains the same.

$$x^{2}\geq \frac{16}{81}$$

Calculate the square root of \frac{16}{81} and get \frac{4}{9}. Rewrite \frac{16}{81} as \left(\frac{4}{9}\right)^{2}.

$$x^{2}\geq \left(\frac{4}{9}\right)^{2}$$

Inequality holds for |x|\geq \frac{4}{9}.

$$|x|\geq \frac{4}{9}$$

Rewrite |x|\geq \frac{4}{9} as x\leq -\frac{4}{9}\text{; }x\geq \frac{4}{9}.

$$x\leq -\frac{4}{9}\text{; }x\geq \frac{4}{9}$$