# Problem 1. (2 points)

Petya came up with a quadratic equation $x^{2}+p x+q$, the roots of which are numbers $x_{1}$ and $x_{2}$. He told Vasya three out of the four numbers $p, q, x_{1}, x_{2}$, without specifying which was which. These turned out to be the numbers $1, 2, -6$. What was the fourth number?
# Answer: -3

Solution:

Notice that by Vieta's theorem $x_{1}+x_{2}+p=0$. However, among the three reported numbers, there are no three that sum up to 0. Therefore, the fourth number must sum to 0 with any two of the existing ones, which means it is either -3, 5, or 4.

Moreover, the condition $q=x_{1} x_{2}$ must be satisfied, where $q$ is the number not included in the zero sum. By checking all possible combinations, it can be confirmed that only $-6=(-3) \cdot 2$ fits.