9.1. Two given quadratic trinomials $f(x)$ and $g(x)$ each have two roots, and the equalities $f(1)=g(2)$ and $g(1)=f(2)$ hold. Find the sum of all four roots of these trinomials.
Answer: 6.

First solution. Let $f(x)=x^{2}+a x+b, g(x)=x^{2}+$ $+c x+d$. Then the conditions of the problem can be written as

$$
1+a+b=4+2 c+d \quad \text { and } \quad 4+2 a+b=1+c+d
$$

Subtracting the second equation from the first, we get $-3-a=3+c$, which means $a+c=-6$. By Vieta's theorem, $-a$ is the sum of the roots of the first quadratic, and $-c$ is the sum of the roots of the second quadratic, from which the required result follows.

Second solution. Consider the auxiliary quadratic polynomial $h(x)=g(3-x)$ (it is also a monic polynomial!). Then $h(x)-f(x)$ is a linear polynomial with roots 1 and 2; hence, it is identically zero, i.e., $f(x)=g(3-x)$. Therefore, if $x_{0}$ is a root of $f(x)$, then $3-x_{0}$ is a root of $g(x)$, and the sum of these two roots is 3. Similarly, the sum of the other roots of these polynomials is also 3.

Comment. Only the correct answer - 0 points.

Only the correct answer with an example of two polynomials satisfying the conditions of the problem - 1 point.

Proved that $a+c=-6$ (in the notation of the first solution) - 3 points.

Proved that $f(x)=g(3-x)$ - 3 points.