Solve 64x^2>28*(1+2x^2)

$$64 { x }^{ 2 } > 28 \times (1+2 { x }^{ 2 } )$$

Solve for x

$$x\in \left(-\infty,-\frac{\sqrt{14}}{2}\right)\cup \left(\frac{\sqrt{14}}{2},\infty\right)$$

Solution Steps

Use the distributive property to multiply 28 by 1+2x^{2}.

$$64x^{2}>28+56x^{2}$$

Subtract 56x^{2} from both sides.

$$64x^{2}-56x^{2}>28$$

Combine 64x^{2} and -56x^{2} to get 8x^{2}.

$$8x^{2}>28$$

Divide both sides by 8. Since 8 is positive, the inequality direction remains the same.

$$x^{2}>\frac{28}{8}$$

Reduce the fraction \frac{28}{8} to lowest terms by extracting and canceling out 4.

$$x^{2}>\frac{7}{2}$$

Calculate the square root of \frac{7}{2} and get \frac{\sqrt{14}}{2}. Rewrite \frac{7}{2} as \left(\frac{\sqrt{14}}{2}\right)^{2}.

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

Inequality holds for |x|>\frac{\sqrt{14}}{2}.

$$|x|>\frac{\sqrt{14}}{2}$$

Rewrite |x|>\frac{\sqrt{14}}{2} as x<-\frac{\sqrt{14}}{2}\text{; }x>\frac{\sqrt{14}}{2}.

$$x<-\frac{\sqrt{14}}{2}\text{; }x>\frac{\sqrt{14}}{2}$$