Answer
Choice A
Work Step by Step
If $k$ is a negative number, such as $k=-1,$ then the equation $x^2=4k
$ becomes
\begin{align*}
x^2&=4(-1)
\\
x^2&=-4
.\end{align*}
Taking the square root of both sides (Square Root Property), the equation above is equivalent to
\begin{align*}
x&=\pm\sqrt{-4}
.\end{align*}
Since the radicand (which is $-4$) in the square root symbol is less than $0,$ then $\sqrt{-4}$ is not a real number. Hence, the given equation (or Choice A) will result to complex solutions if $k$ were a negative number.
