Ans: B
For any real number $x$, $x^2+kx+k+8\ge 0$. Then we have
$\begin{array}{rcl}
\Delta & \le & 0 \\
k^2 -4(1)(k+8) & \le & 0 \\
k^2 -4k -32 & \le & 0 \\
(k-8)(k+4) & \le & 0 \\
\end{array}$
$\therefore -4 \le k \le 8$.
