9.1. The numbers $a$ and $b$ are such that each of the two quadratic trinomials $x^{2} + a x + b$ and $x^{2} + b x + a$ has two distinct roots, and the product of these trinomials has exactly three distinct roots. Find all possible values of the sum of these three roots.

(S. Berlov)
Answer: 0.

Solution: From the condition, it follows that the quadratic polynomials $x^{2}+a x+b$ and $x^{2}+b x+a$ have a common root $x_{0}$, as well as roots $x_{1}$ and $x_{2}$ different from it; in particular, $a \neq b$. Then $0=\left(x_{0}^{2}+a x_{0}+b\right)-\left(x_{0}^{2}+b x_{0}+a\right)=(a-b)\left(x_{0}-1\right)$, that is, $x_{0}=1$. Substituting this root into any quadratic polynomial, we get $0=1+a+b$, that is, $b=-a-1$. Thus, our quadratic polynomials have the form $x^{2}+a x-(a+1)=(x-1)(x+a+1)$ and $x^{2}-(a+1) x+a=(x-1)(x-a)$. Their roots are the numbers $1, a$, and $-(a+1)$ (for $a \neq 1,-2,-1 / 2$ they are distinct), the sum of which is zero.

Remark. After obtaining the equality $0=1+a+b$, the solution can be completed differently. By Vieta's theorem, we have $x_{0}+x_{1}=-a, x_{0}+x_{2}=-b$, from which $x_{0}+x_{1}+x_{2}=-a-b-x_{0}=0$.