Solve x^2-16>0

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

Solve for x

$$x\in \left(-\infty,-4\right)\cup \left(4,\infty\right)$$

Solution Steps

Add 16 to both sides.

$$x^{2}>16$$

Calculate the square root of 16 and get 4. Rewrite 16 as 4^{2}.

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

Inequality holds for |x|>4.

$$|x|>4$$

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

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