Example 3 Given that $a$ is a non-negative integer, the equation $2 x-a \sqrt{1-x}-a+4=0$ has at least one integer root. Then the number of possible values of $a$ is ( ).
(A) 4
(B) 3
(C) 2
(D) 1
(2008, National Junior High School Mathematics League Sichuan Preliminary)
【Analysis】From the problem, we know that the equation about $x$
$$
2 x-a \sqrt{1-x}-a+4=0
$$

has an integer root (let it be $x_{0}$). Then
$$
2 x_{0}-a \sqrt{1-x_{0}}-a+4=0 .
$$

Expressing $x_{0}$ in terms of $a$ is relatively difficult, so consider parameter swapping, expressing $a$ in terms of $x_{0}$. Thus,
$$
\begin{array}{l}
\left(\sqrt{1-x_{0}}+1\right) a=2 x_{0}+4 \\
\Rightarrow a=\frac{2 x_{0}+4}{\sqrt{1-x_{0}}+1} .
\end{array}
$$

By $a \geqslant 0 \Rightarrow 2 x_{0}+4 \geqslant 0 \Rightarrow x_{0} \geqslant-2$.
Also, $1-x_{0} \geqslant 0$, i.e., $x_{0} \leqslant 1$.
Therefore, $-2 \leqslant x_{0} \leqslant 1$.
So, $x_{0}$ can only be $-2,-1,0,1$.
Substituting $-2, 0, 1$ into equation (1) respectively, we get the values of $a$ as $0, 2, 6$. When $x_{0}=-1$, $a$ is not a non-negative integer.
Answer: B.