15. (9) Find $q$, for which $x^{2}+x+q=0$ has two distinct real roots satisfying the relation $x_{1}^{4}+2 x_{1} x_{2}^{2}-x_{2}=19$.
Answer: $q=-3$.

Solution: If $x$ is a root, then $x^{2}=-x-q$, hence $x^{4}=x^{2}+$ $2 q x+q^{2}=(2 q-1) x+q^{2}-q$. Then $x_{1}^{4}+2 x_{1} x_{2}^{2}-x_{2}=(2 q-1) x_{1}+q^{2}-q+$ $2 q x_{2}-x_{2}=(2 q-1)\left(x_{1}+x_{2}\right)+q^{2}-q$. Using Vieta's theorem
$x_{1}+x_{2}=-1$, we get $q^{2}-3 q+1=19$. Solving, we obtain $q_{1}=-3$, $q_{2}=6$, but for $q=6$ the original equation has no roots.

## Lomonosov Moscow State University
## "Conquer Sparrow Hills" Olympiad  Variant 3-1