Solve 12x^2geq10

$$12 x ^ { 2 } \geq 10$$

Solve for x

$$x\in (-\infty,-\frac{\sqrt{30}}{6}]\cup [\frac{\sqrt{30}}{6},\infty)$$

Solution Steps

Divide both sides by 12. Since 12 is positive, the inequality direction remains the same.

$$x^{2}\geq \frac{10}{12}$$

Reduce the fraction \frac{10}{12} to lowest terms by extracting and canceling out 2.

$$x^{2}\geq \frac{5}{6}$$

Calculate the square root of \frac{5}{6} and get \frac{\sqrt{30}}{6}. Rewrite \frac{5}{6} as \left(\frac{\sqrt{30}}{6}\right)^{2}.

$$x^{2}\geq \left(\frac{\sqrt{30}}{6}\right)^{2}$$

Inequality holds for |x|\geq \frac{\sqrt{30}}{6}.

$$|x|\geq \frac{\sqrt{30}}{6}$$

Rewrite |x|\geq \frac{\sqrt{30}}{6} as x\leq -\frac{\sqrt{30}}{6}\text{; }x\geq \frac{\sqrt{30}}{6}.

$$x\leq -\frac{\sqrt{30}}{6}\text{; }x\geq \frac{\sqrt{30}}{6}$$