Solve left(x-2right)^2>left(x+2right)left(x-2right)+8

$${ \left(x-2 \right) }^{ 2 } > \left( x+2 \right) \left( x-2 \right) +8$$

Solve for x

$$x<0$$

Solution Steps

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

$$x^{2}-4x+4>\left(x+2\right)\left(x-2\right)+8$$

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.

$$x^{2}-4x+4>x^{2}-4+8$$

Add -4 and 8 to get 4.

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

Subtract x^{2} from both sides.

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

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

$$-4x+4>4$$

Subtract 4 from both sides.

$$-4x>4-4$$

Subtract 4 from 4 to get 0.

$$-4x>0$$

Product of two numbers is >0 if both are >0 or both are <0. Since -4<0, x must be <0.

$$x<0$$