Example 11 Given real numbers $a, b$ satisfy $3 a^{4}+2 a^{2}-4=0$ and $b^{4}+b^{2}-3=0$. Then $4 a^{-4}+b^{4}=$ ( ).
(A) 7
(B) 8
(C) 9
(D) 10
Solution: Transform the equation $3 a^{4}+2 a^{2}-4=0$ into $\frac{4}{a^{4}}-$ $\frac{2}{a^{2}}-3=0$, and it is known that $-\frac{2}{a^{2}}$ and $b^{2}$ are the two distinct real roots of the equation $x^{2}+x-3=0$.
Let $-\frac{2}{a^{2}}=x_{1}, b^{2}=x_{2}$. By Vieta's formulas, we have $x_{1}+x_{2}=-1, x_{1} x_{2}=-3$.
$\therefore 4 a^{-4}+b^{4}=\left(-\frac{2}{a^{2}}\right)+\left(b^{2}\right)^{2}$ $=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=7$.