2B. If each root of $x^{2}+3 x-7=0$ is increased by the reciprocal of the other root, the resulting roots are those of the equation $2 x^{2}+a x+b=0$. Determine $a$ and $b$.
Solution. Let $\alpha$ and $\beta$ be the roots of $x^{2}+3 x-7=0$. Then for them, $\alpha+\beta=-3$ and $\alpha \beta=-7$. According to the condition, the roots of $2 x^{2}+a x+b=0$ are $\alpha+\frac{1}{\beta}$ and $\beta+\frac{1}{\alpha}$. Then:

$$
\begin{aligned}
& a=-2\left(\alpha+\frac{1}{\beta}+\beta+\frac{1}{\alpha}\right)=2\left(-(-3)-\frac{\alpha+\beta}{\alpha \beta}\right)=2\left(3-\frac{-3}{-7}\right)=\frac{36}{7} \text { and } \\
& b=2\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)=2\left(\alpha \beta+\frac{1}{\alpha \beta}+2\right)=2\left(-7-\frac{1}{7}+2\right)=-\frac{72}{7}
\end{aligned}
$$

3AAB. In the right triangle $A B C$, a point $M$ is chosen on the leg $A C$ such that $\overline{A M}=2 \overline{M C}$. Determine the smallest angle of triangle $A B C$, if $M$ lies on the perpendicular bisector of the hypotenuse $A B$.