Question 8 Find $A^{2}$, where $A$ is the sum of the absolute values of all roots of the equation
$$
x=\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}}
$$

(9th American Invitational Mathematics Examination)
Solve this equation using the "serialization" method, let
$$
\begin{array}{l}
x_{1}=\sqrt{19}+\frac{91}{x}, \\
x_{2}=\sqrt{19}+\frac{91}{x_{1}}, \\
x_{3}=\sqrt{19}+\frac{91}{x_{2}}, \\
x_{4}=\sqrt{19}+\frac{91}{x_{3}}, \\
x=\sqrt{19}+\frac{91}{x_{4}} .
\end{array}
$$
(From (1))
Since $x_{1}, x_{2}, x_{3}, x_{4}, x$ are symmetric (cyclically symmetric), we conjecture that they are all equal.
In fact, from (2) we know they have the same sign (for example, if $x_{1}x_{2} \Rightarrow \stackrel{(3)}{\Rightarrow} xx_{9}^{(6)} x_{3}x_{3} \stackrel{(4)}{\Rightarrow} x_{1}<x_{2}$, which is impossible. Since
$$
x=\sqrt{19}+\frac{91}{x} \Rightarrow x^{2}-\sqrt{19} x-91=0
$$

The roots $x_{1}, x_{2}$ are one positive and one negative, so
$$
\begin{aligned}
A^{2} & =\left(\left|x_{1}\right|+\left|x_{2}\right|\right)^{2}=\left(x_{1}-x_{2}\right)^{2} \\
& =(\sqrt{19})^{2}-4(-91)=383 .
\end{aligned}
$$