Example 5 Given $x_{1}=1, x_{n+1}=\frac{1}{4}\left(5 x_{n}+\sqrt{9 x_{n}^{2}+16}\right)$, find $x_{n}$.
Solve: After rearranging the original recurrence relation and squaring both sides, we get
$$
2 x_{n+1}^{2}-5 x_{n+1} x_{n}+2 x_{n}^{2}-2=0 \text {. }
$$

Replacing $n+1$ with $n-1$ in the above equation, we get
$$
2 x_{n-1}^{2}-5 x_{n-1} x_{n}+2 x_{n}^{2}-2=0 \text {. }
$$
From (1) and (2), we know that $x_{n-1}, x_{n+1}$ are the two roots of the quadratic equation $2 z^{2}-\left(5 x_{n}\right) z+\left(2 x_{n}^{2}-2\right)=0$. Therefore, by Vieta's formulas, we have $x_{n+1}+x_{n-1}=\frac{5}{2} x_{n}(n \geqslant 2)$, and $x_{1}=1, x_{2}=\frac{1}{4}\left(5 x_{1}+\sqrt{9 x_{1}^{2}+16}\right)=\frac{5}{2}$. The characteristic equation of (3) is $r^{2}-\frac{5}{2} r+1=0$, whose roots are $\lambda_{1}=2, \lambda_{2}=\frac{1}{2}$, so $x_{n}=c_{1} \cdot 2^{n}+c_{2}\left(\frac{1}{2}\right)^{n}$.

From $x_{1}=1, x_{2}=\frac{5}{2}$, we get $\left\{\begin{array}{l}2 c_{1}+\frac{1}{2} c_{2}=1, \\ 4 c_{1}+\frac{1}{4} c_{2}=\frac{5}{2},\end{array}\right.$
Solving these, we get $c_{1}=\frac{2}{3}, c_{2}=-\frac{2}{3}$, so, $x_{n}=\frac{2}{3}\left(2^{n}-2^{-n}\right)$.