$4 \cdot 3$ Given the equation $2 x^{2}-9 x+8=0$, find a quadratic equation whose
one root is the reciprocal of the sum of the roots of the original equation, and the other root is the square of the difference of the roots of the original equation.
[Solution] Let $x_{1}, x_{2}$ be the roots of the equation $2 x^{2}-9 x+8=0$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=\frac{9}{2} \\
x_{1} x_{2}=4 .
\end{array}\right.
$$

Let the required equation be $x^{2}+p x+q=0$, with its roots being $x^{\prime}{ }_{1}, x^{\prime}{ }_{2}$. According to the problem, we have
$$
\begin{array}{l}
x_{1}^{\prime}=\frac{1}{x_{1}+x_{2}}=\frac{2}{9}, \\
x^{\prime}{ }_{2}=\left(x_{1}-x_{2}\right)^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}=\frac{81}{4}-16=\frac{17}{4} .
\end{array}
$$

Thus,
$$
\begin{array}{l}
p=-\left(x_{1}^{\prime}+x_{2}^{\prime}\right)=-\frac{161}{36}, \\
q=x_{1}^{\prime} x_{2}^{\prime}=\frac{34}{36},
\end{array}
$$

Therefore, the required equation is $36 x^{2}-161 x+34=0$.