[Solution] The given condition is that the equation $x^{2}+k x+4=0$ has two distinct real roots $x_{1}$, $x_{2}$. Therefore, the discriminant $\Delta=k^{2}-16>0$, or $k^{2}>16$.
By Vieta's formulas, we have $x_{1}+x_{2}=-k, \quad x_{1} x_{2}=4$.
$$
\therefore \quad x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=k^{2}-8>16-8=8 \text {. }
$$

Therefore, the correct choice is (C).
$5 \cdot 17$ Let the parabola $y=x^{2}+k x+4$ intersect the $x$-axis at two different points $\left(x_{1}\right.$, $0),\left(x_{2}, 0\right)$. Then which of the following conclusions must be true?
(A) $x_{1}^{2}+x_{2}^{2}=17$.
(B) $x_{1}^{2}+x_{2}^{2}=8$.
(C) $x_{1}^{2}+x_{2}^{2}>8$.
(D) $x_{1}^{2}+x_{2}^{2}<17$.
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1990)