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

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

Solve for x

$$x>-4$$

Solution Steps

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

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

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

$$x^{2}+4x+4-\left(x^{2}+6x+9\right)<3$$

To find the opposite of x^{2}+6x+9, find the opposite of each term.

$$x^{2}+4x+4-x^{2}-6x-9<3$$

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

$$4x+4-6x-9<3$$

Combine 4x and -6x to get -2x.

$$-2x+4-9<3$$

Subtract 9 from 4 to get -5.

$$-2x-5<3$$

Add 5 to both sides.

$$-2x<3+5$$

Add 3 and 5 to get 8.

$$-2x<8$$

Divide both sides by -2. Since -2 is negative, the inequality direction is changed.

$$x>\frac{8}{-2}$$

Divide 8 by -2 to get -4.

$$x>-4$$