Solve 9x^2>4

$$9 x ^ { 2 } > 4$$

Solve for x

$$x\in \left(-\infty,-\frac{2}{3}\right)\cup \left(\frac{2}{3},\infty\right)$$

Solution Steps

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

$$x^{2}>\frac{4}{9}$$

Calculate the square root of \frac{4}{9} and get \frac{2}{3}. Rewrite \frac{4}{9} as \left(\frac{2}{3}\right)^{2}.

$$x^{2}>\left(\frac{2}{3}\right)^{2}$$

Inequality holds for |x|>\frac{2}{3}.

$$|x|>\frac{2}{3}$$

Rewrite |x|>\frac{2}{3} as x<-\frac{2}{3}\text{; }x>\frac{2}{3}.

$$x<-\frac{2}{3}\text{; }x>\frac{2}{3}$$