4. Given real numbers $\alpha, \beta$ satisfy $\alpha^{2}+3 \alpha-1=0, \beta^{2}$ $-3 \beta-1=0$, and $\alpha \beta \neq 1$. Then the value of $\alpha^{-2}+3 \beta$ is ( ).
(A) 1
(B) 3
(C) -3
(D) 10
4. (D).

From $\alpha^{2}+3 \alpha-1=0$ we know $\alpha \neq 0$ and $\alpha^{-2}-3 \alpha^{-1}-1=0$, meaning $\alpha^{-1}$ and $\beta$ are the two distinct roots of the equation $x^{2}-3 x-1=0$ (since $\alpha \beta \neq 1$), hence
$$
\alpha^{-2}+3 \beta=3 \alpha^{-1}+1+3 \beta=3\left(\alpha^{-1}+\beta\right)+1=10 .
$$