Solve 9k^2geq66

$$9 k ^ { 2 } \geq 66$$

Solve for k

$$k\in (-\infty,-\frac{\sqrt{66}}{3}]\cup [\frac{\sqrt{66}}{3},\infty)$$

Solution Steps

Divide both sides by 9. Since 9 is positive, the inequality direction remains the same.

$$k^{2}\geq \frac{66}{9}$$

Reduce the fraction \frac{66}{9} to lowest terms by extracting and canceling out 3.

$$k^{2}\geq \frac{22}{3}$$

Calculate the square root of \frac{22}{3} and get \frac{\sqrt{66}}{3}. Rewrite \frac{22}{3} as \left(\frac{\sqrt{66}}{3}\right)^{2}.

$$k^{2}\geq \left(\frac{\sqrt{66}}{3}\right)^{2}$$

Inequality holds for |k|\geq \frac{\sqrt{66}}{3}.

$$|k|\geq \frac{\sqrt{66}}{3}$$

Rewrite |k|\geq \frac{\sqrt{66}}{3} as k\leq -\frac{\sqrt{66}}{3}\text{; }k\geq \frac{\sqrt{66}}{3}.

$$k\leq -\frac{\sqrt{66}}{3}\text{; }k\geq \frac{\sqrt{66}}{3}$$