1. $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}$, but according to Vieta's theorem $\left\{\begin{array}{l}D \geq 0 \\ x_{1}+x_{2}=-(m+1) . \text { Then, } \mathrm{c} \\ x_{1} x_{2}=2 m-2\end{array}\right.$

considering that $D=(m+3)^{2} \geq 0$, we have $x_{1}^{2}+x_{2}^{2}=$

$(-(m+1))^{2}-2(2 m-2)=m^{2}+2 m+1-4(m-1)=m^{2}-2 m+5=(m-1)^{2}+4$

From which $y=(m-1)^{2}+4$ and $y_{\text {min }}=4$ when $m=1$.
Answer: For the equation $x^{2}+(m+1) x+2 m-2=0$, the smallest sum of the squares of its roots is 4 when $m=1$.

Grading criteria.

| 15 points | Correct and justified solution. |
| :--- | :--- |
| 10 points | Using Vieta's theorem, the expression for the sum of the squares of the roots is correctly written, but there is an error in the transformation of the expression. |
| 5 points | Vieta's theorem (considering D) is correctly written. |
| 0 points | Other solutions that do not meet the above criteria. |