Solve 1+4/3(m^2-1)>0

$$1 + \frac { 4 } { 3 } ( m ^ { 2 } - 1 ) > 0$$

Solve for m

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

Solution Steps

Use the distributive property to multiply \frac{4}{3} by m^{2}-1.

$$1+\frac{4}{3}m^{2}-\frac{4}{3}>0$$

Subtract \frac{4}{3} from 1 to get -\frac{1}{3}.

$$-\frac{1}{3}+\frac{4}{3}m^{2}>0$$

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

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

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

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

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

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

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

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