Problem 9.1. Find all values of the real parameter $a$ for which the equations $x^{2}-(2 a+1) x+a=0$ and $x^{2}+(a-4) x+a-1=0$ have real roots $x_{1}, x_{2}$ and $x_{3}, x_{4}$, respectively, such that

$$
\frac{x_{1}}{x_{3}}+\frac{x_{4}}{x_{2}}=\frac{x_{1} x_{4}\left(x_{1}+x_{2}+x_{3}+x_{4}\right)}{a}
$$

Peter Boyvalenkov
9.1. For $a \neq 0, a \neq 1$, the given equality is equivalent to

$$
\begin{aligned}
& a\left(x_{1} x_{2}+x_{3} x_{4}\right)=x_{1} x_{2} x_{3} x_{4}\left(x_{1}+x_{2}+x_{3}+x_{4}\right) \\
& \Longleftrightarrow 2 a-1=(a-1)(a+5) \Longleftrightarrow a^{2}+2 a-4=0 \\
& \Longleftrightarrow a_{1,2}=-1 \pm \sqrt{5}
\end{aligned}
$$

It is easy to check that for these values of $a$ both equations have real roots. The case $a=0$ is excluded by the condition and $a=1$ implies $x_{4}=0$, whence $x_{1}=0$, which is a contradiction.