Solve (x+3)^2>(x-2)^2

$$( x + 3 ) ^ { 2 } > ( x - 2 ) ^ { 2 }$$

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}.

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

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

$$x^{2}+6x+9>x^{2}-4x+4$$

Subtract x^{2} from both sides.

$$x^{2}+6x+9-x^{2}>-4x+4$$

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

$$6x+9>-4x+4$$

Add 4x to both sides.

$$6x+9+4x>4$$

Combine 6x and 4x to get 10x.

$$10x+9>4$$

Subtract 9 from both sides.

$$10x>4-9$$

Subtract 9 from 4 to get -5.

$$10x>-5$$

Divide both sides by 10. Since 10 is positive, the inequality direction remains the same.

$$x>\frac{-5}{10}$$

Reduce the fraction \frac{-5}{10} to lowest terms by extracting and canceling out 5.

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