1. Find all values of $a$ for which the quadratic trinomials $x^{2}+2 x+a$ and $x^{2}+a x+2=0$ have two roots each, and the sum of the squares of the roots of the first trinomial is equal to the sum of the squares of the roots of the second trinomial.
Answer: $a=-4$

Solution. If $x_{1}$ and $x_{2}$ are the roots of the quadratic polynomial $x^{2}+p x+q$, then by Vieta's theorem

$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=p^{2}-2 q
$$

Therefore, we need to find such numbers $a$ for which $2^{2}-2 a=a^{2}-2 \cdot 2$. Thus, we need to solve the equation $a^{2}+2 a-8=0$. Its roots are $a=-4$ and $a=2$. The first solution is valid, while the second solution is not valid, since the quadratic polynomial $x^{2}+2 x+2$ does not have roots.