Boodanov I.I.

On the board, nine quadratic trinomials are written: $x^{2}+a_{1} x+b_{1}, x^{2}+a_{2} x+b_{2}, \ldots, x^{2}+a_{9} x+b_{9}$. It is known that the sequences $a_{1}, a_{2}, \ldots, a_{9}$ and $b_{1}, b_{2}, \ldots, b_{9}$ are arithmetic progressions. It turned out that the sum of all nine trinomials has at least one root. What is the maximum number of the original trinomials that can have no roots?
Let $P_{i}(x)=x^{2}+a_{i} x+b_{i}, P(x)=P_{1}(x)+\ldots+P_{9}(x)$. Note that $P_{i}(x)+P_{10-i}(x)=2 P_{5}(x)$. Therefore, $P(x)=$ $9 P_{5}(x)$, and the condition is equivalent to $P_{5}(x)$ having at least one root.

Let $x_{0}$ be one of its roots. Then $P_{i}\left(x_{0}\right)+P_{10-i}\left(x_{0}\right)=2 P_{5}\left(x_{0}\right)$, which means either $P_{i}\left(x_{0}\right) \leq 0$ or $P_{10-i}\left(x_{0}\right) \leq 0$. This implies that in each of the pairs $\left(P_{1}, P_{9}\right),\left(P_{2}, P_{8}\right),\left(P_{3}, P_{7}\right),\left(P_{4}, P_{6}\right)$, at least one of the quadratic polynomials has a root. Therefore, there are at least five quadratic polynomials that have roots, meaning there are no more than four quadratic polynomials without roots.
An example where exactly five quadratic polynomials have at least one root: $x^{2}-4, x^{2}-3, x^{2}-2, \ldots, x^{2}+4$.

## Answer

4 quadratic polynomials.