Answer
\begin{align*}k&=-1 &\text{ or }& &k=1\end{align*}.
Work Step by Step
RECALL:
To complete the square of $x+kx$, we add $\left(\dfrac{k}{2}\right)^2$ to obtain:
$$x^2+kx+\dfrac{k^2}{4}$$
Thus, to find the value of $k$, we simply equate the constant term of the given expressions to $\dfrac{k^2}{4}$.
Hence,
\begin{align*}
\dfrac{k^2}{4}&=\frac{1}{4}\\\\
k^2&=4\left(\frac{1}{4}\right)\\\\
k^2&=1\\\\
k&=\pm\sqrt{1}\\\\
k&=\pm 1
\end{align*}
Thus, $k=-1 \text{ or } k=1$.