Solve (x-1)^2<x(x-4)+8

$$( x - 1 ) ^ { 2 } < x ( x - 4 ) + 8$$

Solve for x

$$x<\frac{7}{2}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x^{2}-2x+1<x\left(x-4\right)+8$$

Use the distributive property to multiply x by x-4.

$$x^{2}-2x+1<x^{2}-4x+8$$

Subtract x^{2} from both sides.

$$x^{2}-2x+1-x^{2}<-4x+8$$

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

$$-2x+1<-4x+8$$

Add 4x to both sides.

$$-2x+1+4x<8$$

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

$$2x+1<8$$

Subtract 1 from both sides.

$$2x<8-1$$

Subtract 1 from 8 to get 7.

$$2x<7$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

$$x<\frac{7}{2}$$