Solve 4-16a^2<0

$$4 - 16 a ^ { 2 } < 0$$

Solve for a

$$a\in \left(-\infty,-\frac{1}{2}\right)\cup \left(\frac{1}{2},\infty\right)$$

Solution Steps

Multiply the inequality by -1 to make the coefficient of the highest power in 4-16a^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-4+16a^{2}>0$$

Add \frac{1}{4} to both sides.

$$a^{2}>\frac{1}{4}$$

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

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

Inequality holds for |a|>\frac{1}{2}.

$$|a|>\frac{1}{2}$$

Rewrite |a|>\frac{1}{2} as a<-\frac{1}{2}\text{; }a>\frac{1}{2}.

$$a<-\frac{1}{2}\text{; }a>\frac{1}{2}$$