2. Given the roots $x_{0}$ and $x_{1}, x_{0}$ and $x_{2}, \ldots, x_{0}$ and $x_{n}$ of the quadratic polynomials

$$
P_{1}(x)=x^{2}+b_{1} x+c_{1}, P_{2}(x)=x^{2}+b_{2} x+c_{2}, \ldots, P_{n}(x)=x^{2}+b_{n} x+c_{n}
$$

respectively. Find the roots of the quadratic polynomial

$$
P(x)=x^{2}+\frac{b_{1}+b_{2}+\ldots+b_{n}}{n} x+\frac{c_{1}+c_{2}+\ldots+c_{n}}{n}
$$
Solution. From the condition of the problem we have

$$
x_{0}^{2}+b_{1} x_{0}+c_{1}=x_{0}^{2}+b_{2} x_{0}+c_{2}=x_{0}^{2}+b_{3} x_{0}+c_{3}=\ldots=x_{0}^{2}+b_{n} x_{0}+c_{n}=0
$$

If we add the last equalities, we get

$$
n x_{0}^{2}+\left(b_{1}+b_{2}+b_{3}+\ldots+b_{n}\right) x_{0}+\left(c_{1}+c_{2}+\ldots+c_{n}\right)=0
$$

i.e., $x_{0}$ is a solution of the polynomial $P(x)$ under consideration. Therefore, the discriminant of $P(x)$ is non-negative and it has a second real root $x^{*}$. From Vieta's formulas, we have:

$$
x_{0}+x_{1}=-b_{1}, x_{0}+x_{2}=-b_{2}, \ldots, x_{0}+x_{n}=-b_{n}, x_{0}+x^{*}=-\frac{1}{n}\left(b_{1}+b_{2}+\ldots+b_{n}\right)
$$

If we multiply the last equality by $n$ and subtract the first $n$ equalities from it, we get $x^{*}=-\frac{1}{n}\left(x_{1}+x_{2}+\ldots+x_{n}\right)$.