Solve (2k-4)^2-4k^2>0

$$( 2 k - 4 ) ^ { 2 } - 4 k ^ { 2 } > 0$$

Solve for k

$$k<1$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2k-4\right)^{2}.

$$4k^{2}-16k+16-4k^{2}>0$$

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

$$-16k+16>0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$-16k>-16$$

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

$$k<\frac{-16}{-16}$$

Divide -16 by -16 to get 1.

$$k<1$$