4. 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.
Given the equation $2 x^{2}-9 x+8=0$ with roots $x_{1}, x_{2}$, then $x_{1}+x_{2}=\frac{9}{2}, x_{1} x_{2}=4$.
Let the roots of the new equation be $x_{1}^{\prime}, x_{2}^{\prime}$, thus $x_{1}^{\prime}=\frac{1}{x_{1}+x_{2}}=\frac{2}{9}$,
$$
x_{2}^{\prime}=\left(x_{1}-x_{2}\right)^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}=\frac{81}{4}-16=\frac{17}{4} \text {. }
$$

Therefore, the new equation is $(9 x-2)(4 x-17)=0 \Rightarrow 36 x^{2}-161 x+34=0$.