Solve 14(1/2x-1)(1/2x+1)-1/2(1-7x^2)>0

$$14 ( \frac { 1 } { 2 } x - 1 ) ( \frac { 1 } { 2 } x + 1 ) - \frac { 1 } { 2 } ( 1 - 7 x ^ { 2 } ) > 0$$

Solve for x

$$x\in \left(-\infty,-\frac{\sqrt{406}}{14}\right)\cup \left(\frac{\sqrt{406}}{14},\infty\right)$$

Solution Steps

Use the distributive property to multiply 14 by \frac{1}{2}x-1.

$$\left(7x-14\right)\left(\frac{1}{2}x+1\right)-\frac{1}{2}\left(1-7x^{2}\right)>0$$

Use the distributive property to multiply 7x-14 by \frac{1}{2}x+1 and combine like terms.

$$\frac{7}{2}x^{2}-14-\frac{1}{2}\left(1-7x^{2}\right)>0$$

Use the distributive property to multiply -\frac{1}{2} by 1-7x^{2}.

$$\frac{7}{2}x^{2}-14-\frac{1}{2}+\frac{7}{2}x^{2}>0$$

Subtract \frac{1}{2} from -14 to get -\frac{29}{2}.

$$\frac{7}{2}x^{2}-\frac{29}{2}+\frac{7}{2}x^{2}>0$$

Combine \frac{7}{2}x^{2} and \frac{7}{2}x^{2} to get 7x^{2}.

$$7x^{2}-\frac{29}{2}>0$$

Add \frac{29}{14} to both sides.

$$x^{2}>\frac{29}{14}$$

Calculate the square root of \frac{29}{14} and get \frac{\sqrt{406}}{14}. Rewrite \frac{29}{14} as \left(\frac{\sqrt{406}}{14}\right)^{2}.

$$x^{2}>\left(\frac{\sqrt{406}}{14}\right)^{2}$$

Inequality holds for |x|>\frac{\sqrt{406}}{14}.

$$|x|>\frac{\sqrt{406}}{14}$$

Rewrite |x|>\frac{\sqrt{406}}{14} as x<-\frac{\sqrt{406}}{14}\text{; }x>\frac{\sqrt{406}}{14}.

$$x<-\frac{\sqrt{406}}{14}\text{; }x>\frac{\sqrt{406}}{14}$$