* 5. For the quartic equation $x^{4}-20 x^{3} + k x^{2}-400 x+384=0$, the product of two of its four roots is 24, then the value of $k$ is $\qquad$.
5. 140 .

Let the four roots of the equation be $x_{1}, x_{2}, x_{3}, x_{4}$, then by Vieta's formulas, we have
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}=20, \\
x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4}=k \text { (2) } \\
x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4}=400, \\
x_{1} x_{2} x_{3} x_{4}=384 .
\end{array}\right.
$$

Without loss of generality, let $x_{1} x_{2}=24$.
From (4), we get $x_{3} x_{4}=16$.
Substituting (5) and (3) into (2) and (3), we get
$$
\begin{array}{l}
24+\left(x_{1}+x_{2}\right)\left(x_{3}+x_{4}\right)+16=k, \\
24\left(x_{3}+x_{4}\right)+16\left(x_{1}+x_{2}\right)=400 .
\end{array}
$$

Solving (1) and (8) simultaneously, we get $x_{3}+x_{4}=10, x_{1}+x_{2}=10$.
Substituting into (7), we get $k=24+10 \cdot 10+16=140$.