Example 5 If 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
$$
\begin{array}{l} 
\text { From the problem, we know that } x^{4}+a x^{2}+b x+c \text { can definitely be divided by } x^{2}-3 x-1. \\
=\left(x^{2}-3 x-1\right)\left(x^{2}+3 x+a+10\right)+ \\
{[(3 a+b+33) x+(a+c+10)], } \\
\text { then }\left\{\begin{array}{l}
3 a+b+33=0, \\
a+c+10=0
\end{array}\right. \\
\Rightarrow a+b-2 c=-13 .
\end{array}
$$

Therefore, the answer is A.