Example 3 Let $a b \neq 0$, and the functions
$$
\begin{aligned}
f_{1}(x) & =x^{2}+2 a x+4 b \\
\text { and } f_{2}(x) & =x^{2}+4 a x+2 b
\end{aligned}
$$

have the same minimum value $u$, and the functions
$$
\begin{aligned}
f_{3}(x) & =-x^{2}+2 b x+4 a \\
\text { and } f_{4}(x) & =-x^{2}+4 b x+2 a
\end{aligned}
$$

have the same maximum value $v$. Then the value of $u+v$ ( ).
(A) must be a positive number
(B) must be a negative number
(C) must be 0
(D) the sign cannot be determined ${ }^{[3]}$
Notice that,
$$
\begin{array}{l}
f_{1}(x)=(x+a)^{2}+4 b-a^{2} \geqslant 4 b-a^{2}, \\
f_{2}(x)=(x+2 a)^{2}+2 b-4 a^{2} \geqslant 2 b-4 a^{2}, \\
f_{3}(x)=-(x-b)^{2}+4 a+b^{2} \leqslant 4 a+b^{2}, \\
f_{4}(x)=-(x-2 b)^{2}+2 a+4 b^{2} \leqslant 2 a+4 b^{2} .
\end{array}
$$

From $4 b-a^{2}=u=2 b-4 a^{2} \Rightarrow-2 b=3 a^{2}$;
From $4 a+b^{2}=v=2 a+4 b^{2} \Rightarrow 2 a=3 b^{2}$.
(1) $\times$ (2) gives $-4 a b=9 a^{2} b^{2} \Rightarrow a b=-\frac{4}{9}$.

Substituting into equations (1) and (2) yields $a=\frac{2}{3}, b=-\frac{2}{3}$.
Thus, $u+v=0$, so the answer is C.