Three. (20 points) How many real roots does the equation $\sqrt{x} \sin x^{2}-2=0$ have in the interval $[0,20]$?
Three, Solution 1: Let $y=\sqrt{x} \sin x^{2}$.
The solutions to the equation $y=0$ are $x=\sqrt{k \pi} \in[0,20], k \in \mathbf{Z}$. Therefore, $x=\sqrt{k \pi}(k=0,1, \cdots, 127)$.
When $\sqrt{x} \geqslant 2$, the graph of $y=\sqrt{x} \sin x^{2}$ may intersect with the graph of $y=2$, so $x \geqslant 4$.

Also, $\sqrt{5 \pi}4$, when $x \in[\sqrt{6 \pi}, \sqrt{7 \pi}]$, $\sqrt{x} \sin x^{2} \geqslant 0$, the function $y=\sqrt{x} \sin x^{2}$ increases first and then decreases on $[\sqrt{6 \pi}, \sqrt{7 \pi}]$, and when $x=\sqrt{6 \pi+\frac{\pi}{2}}$,
$$
\begin{array}{l}
y_{\max } \geqslant \sqrt[4]{6 \pi+\frac{\pi}{2}} \sin \left(\sqrt{6 \pi+\frac{\pi}{2}}\right)^{2} \\
=\sqrt[4]{6 \pi+\frac{\pi}{2}}>2 .
\end{array}
$$

Therefore, within the interval $[\sqrt{6 \pi}, \sqrt{7 \pi}]$, the equation $\sqrt{x} \sin x^{2}-2=0$ has two real roots.

Similarly, within the intervals $[\sqrt{8 \pi}, \sqrt{9 \pi}],[\sqrt{10 \pi}, \sqrt{11 \pi}], \cdots$, $[\sqrt{126 \pi}, \sqrt{127 \pi}]$, the equation $\sqrt{x} \sin x^{2}-2=0$ has two real roots each.
Thus, there are a total of 122 real roots.
In summary, within the interval $[0,20]$, the equation $\sqrt{x} \sin x^{2}-2=0$ has a total of 122 real roots.

Solution 2: Use a graphing calculator to plot the graph of the function $y=\sqrt{x} \sin x^{2}$. To clearly see the number of roots, consider observing several intervals.

For example, when $x \in[0,10]$, i.e., $x_{\text {min }}=0, x_{\max }=10$, the equation has 26 real roots.
Similarly, when $x \in(10,13]$, the equation has 22 real roots;
when $x \in(13,17]$, the equation has 38 real roots;
when $x \in(17,20]$, the equation has 36 real roots.
Therefore, within the interval $[0,20]$, the equation $\sqrt{x} \sin x^{2}-2=0$ has a total of 122 real roots.