答案：A
$\begin{array}{rcl}
x^2 +2x – k & = & 2 \\
x^2 +2x +(-k -2) & = & 0
\end{array}$
$\begin{array}{rcl}
x^2 +2x – k & = & 2 \\
x^2 +2x +(-k -2) & = & 0
\end{array}$
由於以上的二次方程有兩個相異實根，可得
$\begin{array}{rcl}
\Delta & > & 0 \\
(2)^2 -4(1)(-k-2) & > & 0 \\
4 + 4k + 8 & > & 0 \\
4k & > & -12 \\
k & > & -3
\end{array}$
