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.,
$$
x_{1} x_{2}=-32 \text {. }
$$

From (4) $\div$ (5) we get
$$
x_{3} x_{4}=62 \text {. }
$$

Substituting (3) 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 two unknowns, solving (1) and (7) simultaneously we get
$$
x_{1}+x_{2}=4, x_{3}+x_{4}=14 \text {, }
$$

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