# 8. Variant 1.

Consider the quadratic trinomial $P(x)=a x^{2}+b x+c$, which has distinct positive roots. Vasya wrote four numbers on the board: the roots of $P(x)$, as well as the roots of the trinomial $Q(x)=c x^{2}+b x+a$ multiplied by 4. What is the smallest integer value that the sum of the written numbers can have
Answer: 9.

Solution.

Let the roots of the quadratic polynomial $P(x)$ be $x_{1}, x_{2}>0$. Then, by Vieta's formulas, the roots of the polynomial $Q(x)$ will be $\frac{1}{x_{1}}, \frac{1}{x_{2}}$.

The sum of the numbers written on the board is $S=x_{1}+x_{2}+\frac{4}{x_{1}}+\frac{4}{x_{2}}$. For a positive $k$, the inequality $k+\frac{4}{k} \geq 4$ holds, and equality is achieved if and only if $k=2$.

Therefore, $S \geq 8$. However, $S=8$ only if $x_{1}=x_{2}=2$. But according to the problem, $x_{1}, x_{2}$ are distinct. Therefore, $S>8$. Thus, the smallest possible integer value of $S$ is 9. This situation is possible. Let $x_{1}=1$, and $x_{2}$ be such that $x_{2}+\frac{4}{x_{2}}=5$. Then the quadratic polynomial with roots $x_{1}, x_{2}$ is the desired one.