Solve left(x-3right)^2<left(x+4right)^2

$${ \left(x-3 \right) }^{ 2 } < { \left(x+4 \right) }^{ 2 }$$

Solve for x

$$x>-\frac{1}{2}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9<\left(x+4\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$x^{2}-6x+9<x^{2}+8x+16$$

Subtract x^{2} from both sides.

$$x^{2}-6x+9-x^{2}<8x+16$$

Combine x^{2} and -x^{2} to get 0.

$$-6x+9<8x+16$$

Subtract 8x from both sides.

$$-6x+9-8x<16$$

Combine -6x and -8x to get -14x.

$$-14x+9<16$$

Subtract 9 from both sides.

$$-14x<16-9$$

Subtract 9 from 16 to get 7.

$$-14x<7$$

Divide both sides by -14. Since -14 is negative, the inequality direction is changed.

$$x>\frac{7}{-14}$$

Reduce the fraction \frac{7}{-14} to lowest terms by extracting and canceling out 7.

$$x>-\frac{1}{2}$$