4A. Let $x_{0}$ and $x_{1}$ be real roots of the equation $x^{2}+a_{1} x+b_{1}=0$, $x_{0}$ and $x_{2}$ be real roots of the equation $x^{2}+a_{2} x+b_{2}=0$ and $x_{0}$ and $x_{3}$ be real roots of the equation $x^{2}+a_{3} x+b_{3}=0$. Find the roots of the equation

$$
x^{2}+\frac{a_{1}+a_{2}+a_{3}}{3} x+\frac{b_{1}+b_{2}+b_{3}}{3}=0
$$
Solution. From $x_{0}^{2}+a_{1} x_{0}+b_{1}=0, x_{0}^{2}+a_{2} x_{0}+b_{2}=0$ and $x_{0}^{2}+a_{3} x_{0}+b_{3}=0$ we obtain

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

or equivalently

$$
x_{0}^{2}+\frac{a_{1}+a_{2}+a_{3}}{3} x_{0}+\frac{b_{1}+b_{2}+b_{3}}{3}=0
$$

so $x_{0}$ is a real root of the given equation. Then, the equation has another real root. From $x_{0}+x_{1}=-a_{1}, x_{0}+x_{2}=-a_{2}$ and $x_{0}+x_{3}=-a_{3}$ we obtain

$$
3 x_{0}+x_{1}+x_{2}+x_{3}=-\left(a_{1}+a_{2}+a_{3}\right)
$$

or equivalently

$$
x_{0}+\frac{x_{1}+x_{2}+x_{3}}{3}=-\frac{a_{1}+a_{2}+a_{3}}{3}
$$

Since

$$
x_{0} \cdot \frac{x_{1}+x_{2}+x_{3}}{3}=\frac{x_{0} x_{1}+x_{0} x_{2}+x_{0} x_{3}}{3}=\frac{b_{1}+b_{2}+b_{3}}{3}
$$

it follows that $x_{0}$ and $\frac{x_{1}+x_{2}+x_{3}}{3}$ are roots of the given equation.