2. Let the two roots of the equation $x^{2}+x-1=0$ be $\alpha, \beta$. Then the linear equation with roots $\alpha^{2}+\beta^{2}$ and $(\alpha+\beta)^{2}$ is ( ).
(A) $x^{2}+4 x-3=0$
(B) $x^{2}-4 x+3=0$
(C.) $x^{2}+4 x-3=0$
(D) $x^{2}-4 x-3=0$
2. (B).
$$
\begin{array}{l}
\alpha+\beta=-1, \alpha \beta=-1 \Rightarrow(\alpha+\beta)^{2}=1 . \\
\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=1-2(-1)=3 .
\end{array}
$$

Therefore, the equation with roots 1 and 3 is $x^{2}-4 x+3=0$.