Solve 16x^2-1>0

$$16 { x }^{ 2 } -1 > 0$$

Solve for x

$$x\in \left(-\infty,-\frac{1}{4}\right)\cup \left(\frac{1}{4},\infty\right)$$

Solution Steps

Add \frac{1}{16} to both sides.

$$x^{2}>\frac{1}{16}$$

Calculate the square root of \frac{1}{16} and get \frac{1}{4}. Rewrite \frac{1}{16} as \left(\frac{1}{4}\right)^{2}.

$$x^{2}>\left(\frac{1}{4}\right)^{2}$$

Inequality holds for |x|>\frac{1}{4}.

$$|x|>\frac{1}{4}$$

Rewrite |x|>\frac{1}{4} as x<-\frac{1}{4}\text{; }x>\frac{1}{4}.

$$x<-\frac{1}{4}\text{; }x>\frac{1}{4}$$