Solve 8a^2-16>0

$$8 a ^ { 2 } - 16 > 0$$

Solve for a

$$a\in \left(-\infty,-\sqrt{2}\right)\cup \left(\sqrt{2},\infty\right)$$

Solution Steps

Add 2 to both sides.

$$a^{2}>2$$

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

$$a^{2}>\left(\sqrt{2}\right)^{2}$$

Inequality holds for |a|>\sqrt{2}.

$$|a|>\sqrt{2}$$

Rewrite |a|>\sqrt{2} as a<-\sqrt{2}\text{; }a>\sqrt{2}.

$$a<-\sqrt{2}\text{; }a>\sqrt{2}$$