Ans: C
$\begin{array}{rcl}
(x-k)^2 & = & 4k^2 \\
(x-k)^2 – 4k^2 & = & 0 \\
(x-k)^2 – (2k)^2 & = & 0 \\
(x-k+2k)(x-k-2k) & = & 0 \\
(x+k)(x-3k) & = & 0
\end{array}$
$\begin{array}{rcl}
(x-k)^2 & = & 4k^2 \\
(x-k)^2 – 4k^2 & = & 0 \\
(x-k)^2 – (2k)^2 & = & 0 \\
(x-k+2k)(x-k-2k) & = & 0 \\
(x+k)(x-3k) & = & 0
\end{array}$
Therefore, $x=-k$ or $x=3k$.
