Solve left(2x-8right)^2-4(x-2)(x-4)>0

$${ \left(2x-8 \right) }^{ 2 } -4(x-2)(x-4) > 0$$

Solve for x

$$x<4$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-8\right)^{2}.

$$4x^{2}-32x+64-4\left(x-2\right)\left(x-4\right)>0$$

Use the distributive property to multiply -4 by x-2.

$$4x^{2}-32x+64+\left(-4x+8\right)\left(x-4\right)>0$$

Use the distributive property to multiply -4x+8 by x-4 and combine like terms.

$$4x^{2}-32x+64-4x^{2}+24x-32>0$$

Combine 4x^{2} and -4x^{2} to get 0.

$$-32x+64+24x-32>0$$

Combine -32x and 24x to get -8x.

$$-8x+64-32>0$$

Subtract 32 from 64 to get 32.

$$-8x+32>0$$

Subtract 32 from both sides. Anything subtracted from zero gives its negation.

$$-8x>-32$$

Divide both sides by -8. Since -8 is negative, the inequality direction is changed.

$$x<\frac{-32}{-8}$$

Divide -32 by -8 to get 4.

$$x<4$$