4. Given that the two roots of the equation $x^{2}-3 x-1=0$ are also roots of the equation $x^{4}+a x^{2}+b x+c=0$. Then the value of $a+b-2 c$ is ( ).
(A) -13
(B) -9
(C) 6
(D) 0
4. A.

Let $m$ be a root of the equation $x^{2}-3 x-1=0$. Then $m^{2}=3 m+1$.
Given that $m^{4}+a m^{2}+b m+c=0$.
Substituting equation (1) into the above equation and simplifying, we get
$$
(9+a) m^{2}+(6+b) m+c+1=0 \text {. }
$$

Thus, the roots of the equation $x^{2}-3 x-1=0$ are also the roots of the equation $(9+a) x^{2}+(6+b) x+c+1=0$. These two equations are essentially the same quadratic equation.
Therefore, $\frac{9+a}{1}=\frac{6+b}{-3}=\frac{c+1}{-1}$.
So, $b=-3 a-33, c=-a-10$.
Hence, $a+b-2 c=-13$.