2. If the roots of the equation $(x-a)(x-b)=M$ are $\alpha, 3$, then, the sum of the squares of the roots of the equation $(x-\alpha)(x-\beta)=-M$ is ).
(A) $a^{2}+b$
(B) $a+b^{2}$
(C) $a^{2}+b+b^{2}$
(D) $a^{2}+b^{2}$
2. (D):
$x_{1} x_{2}=M+2 \beta, x_{1}+x_{2}=\alpha+\beta$, thus.
$$
\begin{aligned}
x_{1}^{2}+x_{2}^{2} & =\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2} \\
& =(\alpha+\beta)^{2}-2(M+2 \beta) .
\end{aligned}
$$

And $a+\beta=a+b, \alpha \beta=a b-M$.
Therefore, $x_{1}^{2}+x_{2}^{2}=a^{2}+b^{2}$.