## Task B-2.1.

In the equation $x^{2}+m-3 x=m x-2$, determine the positive real number $m$ so that the total sum of all solutions of the equation and their squares is 44.
## Solution.

Given the equation $x^{2}+m-3 x=m x-2$ we write it in the form $x^{2}-(m+3) x+m+2=0. \quad 1$ point The solutions of this quadratic equation must satisfy the condition $x_{1}+x_{2}+x_{1}^{2}+x_{2}^{2}=44$, or $x_{1}+x_{2}+\left(x_{1}+x_{2}\right)^{2}-2 x_{1} \cdot x_{2}=44$.

According to Vieta's formulas, we have

$$
\begin{array}{r}
x_{1}+x_{2}=m+3 \\
x_{1} x_{2}=m+2
\end{array}
$$

Substituting into the given condition $m+3+(m+3)^{2}-2(m+2)=44$ and after simplification, we get $m^{2}+5 m-36=0$.

This equation has solutions $m_{1}=-9, m_{2}=4$.

Since a positive number $m$ is sought, the solution is $m=4$.