Solve [1/4(x+2)^2-4-x]4<x^2-2x

$$[ \frac { 1 } { 4 } ( x + 2 ) ^ { 2 } - 4 - x ] 4 < x ^ { 2 } - 2 x$$

Solve for x

$$x<6$$

Solution Steps

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

$$\left(\frac{1}{4}\left(x^{2}+4x+4\right)-4-x\right)\times 4<x^{2}-2x$$

Use the distributive property to multiply \frac{1}{4} by x^{2}+4x+4.

$$\left(\frac{1}{4}x^{2}+x+1-4-x\right)\times 4<x^{2}-2x$$

Subtract 4 from 1 to get -3.

$$\left(\frac{1}{4}x^{2}+x-3-x\right)\times 4<x^{2}-2x$$

Combine x and -x to get 0.

$$\left(\frac{1}{4}x^{2}-3\right)\times 4<x^{2}-2x$$

Use the distributive property to multiply \frac{1}{4}x^{2}-3 by 4.

$$x^{2}-12<x^{2}-2x$$

Subtract x^{2} from both sides.

$$x^{2}-12-x^{2}<-2x$$

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

$$-12<-2x$$

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

$$-2x>-12$$

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

$$x<\frac{-12}{-2}$$

Divide -12 by -2 to get 6.

$$x<6$$