10.2. 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?

(I. Bogdanov)
10.2. Answer. 4.

Let $P_{i}(x)=x^{2}+a_{i} x+b_{i}, P(x)=P_{1}(x)+\ldots+P_{9}(x)$. Notice that $P_{i}(x)+P_{10-i}(x)=2 x^{2}+\left(a_{i}+a_{10-i}\right) x+\left(b_{i}+b_{10-i}\right)=$ $=2 P_{5}(x)$. 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 any of its roots. Then $P_{i}\left(x_{0}\right)+P_{10-i}\left(x_{0}\right)=2 P_{5}\left(x_{0}\right)=0$, which means that either $P_{i}\left(x_{0}\right) \leqslant 0$ or $P_{10-i}\left(x_{0}\right) \leqslant 0$. Since the leading coefficients of the quadratic polynomials are positive, this implies that in each of the pairs $(P_{1}, P_{9})$, $\left(P_{2}, P_{8}\right),\left(P_{3}, P_{7}\right),\left(P_{4}, P_{6}\right)$, at least one of the polynomials has a root. Therefore, there are at least five polynomials that have at least one root. Hence, there are no more than four polynomials without roots.

It remains to provide an example where exactly five polynomials (one of them being $-P_{5}$) have at least one root. For example, the polynomials $x^{2}-4, x^{2}-3, x^{2}-2, \ldots, x^{2}+4$ will do.