1. The equation $x^{2}+2 a x+2 a^{2}+4 a+3=0$ has real solutions. For which value of the real parameter $a$, is the sum of the squares of its solutions the largest?
Solution. The equation has solutions if $D=4 a^{2}-4\left(2 a^{2}+4 a+3\right) \geq 0$. We get $a^{2}+4 a+3 \leq 0$, that is, $(a+1)(a+3) \leq 0$ and hence $a \in[-3,-1]$. If we denote the solutions of the equation by $x_{1}$ and $x_{2}$, for the sum of their squares we have $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2} \cdot$ Since

$$
x_{1}+x_{2}=-2 a \text { and } x_{1} x_{2}=2 a^{2}+4 a+3
$$

we get

$$
x_{1}^{2}+x_{2}^{2}=4 a^{2}-4 a^{2}-8 a-6=-8 a-6
$$

Then the maximum value of $x_{1}^{2}+x_{2}^{2}$ is achieved for $a=-3$, and it is 18.