4・60 Quartic Equation
$$x^{4}-18 x^{3}+k x^{2}+200 x-1984=0$$

The product of two of the four roots of the equation is -32. Determine the value of $k$.
[Solution] Let the four roots of the equation be $x_{1}, x_{2}, x_{3}, x_{4}$. Then, by Vieta's formulas, we have
$$\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}=18 \\
x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4}=k \\
x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4}=-200 \\
x_{1} x_{2} x_{3} x_{4}=-1984
\end{array}$$

Suppose the product of $x_{1}$ and $x_{2}$ is -32, i.e.,
$$x_{1} x_{2}=-32$$

By (4) $\div$ (5), we get
$$x_{3} x_{4}=62$$

Substituting (5) and (6) into (3) and (2), we get
$$\begin{array}{l}
-32\left(x_{3}+x_{4}\right)+62\left(x_{1}+x_{2}\right)=-200 \\
-32+\left(x_{1}+x_{2}\right)\left(x_{3}+x_{4}\right)+62=k
\end{array}$$

Considering $x_{1}+x_{2}$ and $x_{3}+x_{4}$ as unknowns, solving (1) and (7) together, we get
$$x_{1}+x_{2}=4, x_{3}+x_{4}=14$$

Substituting into (8), we get
$$k=-32+56+62=86 .$$