9.1. For which $p$ and $q$ does the equation $x^{2}+p x+q=0$ have two distinct solutions $2 p$ and $p+q$? Answer: $p=2 / 3, q=-8 / 3$.
Solution. By Vieta's theorem $3 p+q=-p, 2 p(p+q)=q$. From the first equation $q=-4 p$. Substituting into the second, we get $6 p^{2}=4 p$, from which $p=0$ or $p=2 / 3$. The first case does not satisfy the condition.