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

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

Solve for x

$$x<\frac{13}{6}$$

Solution Steps

Consider \left(x+2\right)\left(x-2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

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

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

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

Subtract x^{2} from both sides.

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

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

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

Swap sides so that all variable terms are on the left hand side. This changes the sign direction.

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

Subtract 9 from both sides.

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

Subtract 9 from -4 to get -13.

$$-6x>-13$$

Divide both sides by -6. Since -6 is negative, the inequality direction is changed.

$$x<\frac{-13}{-6}$$

Fraction \frac{-13}{-6} can be simplified to \frac{13}{6} by removing the negative sign from both the numerator and the denominator.

$$x<\frac{13}{6}$$