1B. In the quadratic equation $x^{2}-(m+1) x+3 m+2=0$, determine the parameter $m$ so that the sum of the solutions of the given equation is equal to the sum of their squares.
Solution. By Vieta's formulas, we have $x_{1}+x_{2}=m+1$ and $x_{1} x_{2}=3 m+2$. From the condition of the problem, we have

$$
m+1=x_{1}+x_{2}=x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(m+1)^{2}-2(3 m+2)
$$

i.e., $m^{2}-5 m-4=0$. (10) By solving the quadratic equation, we get $m_{1}=\frac{5-\sqrt{31}}{2}$ and $m_{2}=\frac{5+\sqrt{31}}{2}$.