Solve (x-2)^2>x(x+12)

$$( x - 2 ) ^ { 2 } > x ( x + 12 )$$

Solve for x

$$x<\frac{1}{4}$$

Solution Steps

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

$$x^{2}-4x+4>x\left(x+12\right)$$

Use the distributive property to multiply x by x+12.

$$x^{2}-4x+4>x^{2}+12x$$

Subtract x^{2} from both sides.

$$x^{2}-4x+4-x^{2}>12x$$

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

$$-4x+4>12x$$

Subtract 12x from both sides.

$$-4x+4-12x>0$$

Combine -4x and -12x to get -16x.

$$-16x+4>0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$-16x>-4$$

Divide both sides by -16. Since -16 is negative, the inequality direction is changed.

$$x<\frac{-4}{-16}$$

Reduce the fraction \frac{-4}{-16} to lowest terms by extracting and canceling out -4.

$$x<\frac{1}{4}$$