Example 3 (1995 National High School League Question) Given the equation $|x-2 n|=k \sqrt{x}(n \in \mathbf{N})$ has two distinct real roots in the interval $(2 n-$ $1,2 n+1]$, then the range of values for $k$ is ( ).
A. $k>0$
B. $0<k \leqslant \frac{1}{\sqrt{2 n+1}}$
C. $\frac{1}{2 n+1}<k \leqslant \frac{1}{\sqrt{2 n+1}}$
D. None of the above
Solution: Choose B. Reason: Clearly $k \geqslant 0$, and $k=0$ leads to $x=2 n$, on the interval $(2 n-1,2 n+1]$ the original equation has only one root, so $k>0$.

From $(x-2 n)^{2}=k^{2} x$, we know that the parabola $y=(x-2 n)^{2}$ and the line $y=k^{2} x$ have two different intersection points on the interval $(2 n-1, 2 n+1]$. Therefore, when $x=2 n-1$, we have $(x-2 n)^{2}>k^{2} x$, and when $x=2 n+1$, we have $(x-2 n)^{2} \geqslant k^{2} x$, thus $k^{2}(2 n+1) \leqslant 1$, i.e., $k \leqslant \frac{1}{\sqrt{2 n+1}}$.