For example. Solve the equation: $\sqrt{x+3}+\sqrt{2 x-1}=4$.
Solve for $(\sqrt{x+3}+\sqrt{2 x-1})^{2}$
$$
\begin{array}{c}
=\left(\sqrt{2} \cdot \sqrt{x+3}+\frac{1}{\sqrt{2}}\right. \\
\cdot \sqrt{2 x-1})^{2}+(1-2) \cdot\left(3+\frac{1}{2}\right),
\end{array}
$$

i.e., $16=\left(\sqrt{2} \cdot \sqrt{x+3}+\frac{1}{\sqrt{2}} \sqrt{2 x-1}\right)^{2}$
$$
-\frac{7}{2} \text {, }
$$

The corresponding system of equations is
$$
\left\{\begin{array}{l}
\sqrt{x+3}+\sqrt{2 x-1}=4, \\
2 \sqrt{x+3}+\sqrt{2 x-1}=\sqrt{39} .
\end{array}\right.
$$

Solving by elimination, we get
$$
\begin{aligned}
\sqrt{x+3}=\sqrt{39}-4 \Rightarrow x=52-8 \sqrt{39}, \\
\sqrt{2 x-1}=8-\sqrt{39} \Rightarrow x=52-8 \sqrt{39} . \\
\text { The original equation has only one root } x=52-8 \sqrt{39} .
\end{aligned}
$$
$\therefore$ The original equation has only one root $x=52-8 \sqrt{39}$.
$$
\text { 2. } \sqrt{f(x)} \pm \sqrt{g(x)}=m
$$

A type of solution for equations of the form $(f(x)$ and $g(x)$ are rational expressions of $x$, and $m \neq 0$ is a constant)
Clearly, $[\sqrt{f(x)}]^{2}-[\sqrt{g(x)})^{2}$
$$
=f(x)-g(x) \text {. }
$$
(2) gives $\sqrt{f(x)} \mp \sqrt{g(x)}=\frac{f(x)-g(x)}{m}$.

i.e., $m \sqrt{f(x)} \pm m \sqrt{g(x)}$
$$
=f(x)-g(x) \text {. }
$$
(1) $\times m+$ (3), we get $2 m \sqrt{f(x)}=f(x)-g(x)+m^{2}$.
$$
\text { Hence } 4 m^{2} \cdot f(x)=\left[f(x)-g(x)+m^{2}\right]^{2} \text {.(4) }
$$

It can be proven that if $x=a$ is a solution to equation (1), then it must also be a solution to equation (4) (necessary condition). Therefore, to solve the irrational equation (1), it is only necessary to solve the rational equation (4). Then, substitute the obtained solutions into (1) for verification.