### 6.345 Find the coefficients $p$ and $q$ of the equation

$x^{4}-10 x^{3}+37 x^{2}+p x+q=0$, if among its roots there are two pairs of equal numbers.
Solution.

Let $x_{1}=x_{2}, x_{3}=x_{4}$ be the roots of the original equation. By Vieta's theorem, we have:

$\left\{\begin{array}{l}x_{1}+x_{3}=5, \\ x_{1}^{2}+4 x_{1} x_{3}+x_{3}^{2}=37, \\ 2 x_{1} x_{3}\left(x_{1}+x_{3}\right)=-\dot{p}, \\ \left(x_{1} x_{3}\right)^{2}=q\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{1}+x_{3}=5, \\ \left(x_{1}+x_{3}\right)^{2}+2 x_{1} x_{3}=37, \\ 2 x_{1} x_{3}\left(x_{1}+x_{3}\right)=-p, \\ \left(x_{1} x_{3}\right)^{2}=q\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{1} x_{3}=6, \\ 10 x_{1} x_{3}=-p, \\ \left(x_{1} x_{3}\right)^{2}=q\end{array} \Rightarrow\right.\right.\right.$

$\Rightarrow p=-60, q=36$.

Answer: $p=-60, q=36$.