Solve 400-128a^2+256<0

$$400-128 { a }^{ 2 } +256 < 0$$

Solve for a

$$a\in \left(-\infty,-\frac{\sqrt{82}}{4}\right)\cup \left(\frac{\sqrt{82}}{4},\infty\right)$$

Solution Steps

Add 400 and 256 to get 656.

$$656-128a^{2}<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 656-128a^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-656+128a^{2}>0$$

Add \frac{41}{8} to both sides.

$$a^{2}>\frac{41}{8}$$

Calculate the square root of \frac{41}{8} and get \frac{\sqrt{82}}{4}. Rewrite \frac{41}{8} as \left(\frac{\sqrt{82}}{4}\right)^{2}.

$$a^{2}>\left(\frac{\sqrt{82}}{4}\right)^{2}$$

Inequality holds for |a|>\frac{\sqrt{82}}{4}.

$$|a|>\frac{\sqrt{82}}{4}$$

Rewrite |a|>\frac{\sqrt{82}}{4} as a<-\frac{\sqrt{82}}{4}\text{; }a>\frac{\sqrt{82}}{4}.

$$a<-\frac{\sqrt{82}}{4}\text{; }a>\frac{\sqrt{82}}{4}$$