2. The quadratic trinomial $x^{2}+u x-v$ has distinct non-zero roots $p$ and $q$, and the quadratic trinomial $x^{2}+p x-q-$ has distinct non-zero roots $u$ and $v$. Find all possible values of $p, q, u$ and $v$.
Answer. $p=-1, q=2, u=-1, v=2$.

Solution. Applying Vieta's theorem to both equations, we form the system of equations:

$$
\left\{\begin{array}{l}
p+q=-u \\
u+v=-p \\
p q=-v \\
u v=-q
\end{array}\right.
$$

Subtracting the second equation from the first, after simplification, we get: $q=v$. Substituting $q$ for $v$ in the third and fourth equations and simplifying by $q \neq 0$, we find that $p=u=-1$. Now, from the first two equations, we find $q=v=2$.