Solve (k+1)^2-4(1/4k^2)>0

$$( k + 1 ) ^ { 2 } - 4 ( \frac { 1 } { 4 } k ^ { 2 } ) > 0$$

Solve for k

$$k>-\frac{1}{2}$$

Solution Steps

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

$$k^{2}+2k+1-4\times \frac{1}{4}k^{2}>0$$

Multiply 4 and \frac{1}{4} to get 1.

$$k^{2}+2k+1-k^{2}>0$$

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

$$2k+1>0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$2k>-1$$

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

$$k>-\frac{1}{2}$$