8.50 For $x \in[0,2 \pi]$, and satisfying the equation $\operatorname{tg}^{2} x-9 \operatorname{tg} x+1=0$, the sum of all roots is
(A) $\frac{\pi}{2}$.
(B) $\pi$.
(C) $\frac{3 \pi}{2}$.
(D) $3 \pi$.
(E) $4 \pi$.
(40th American High School Mathematics Examination, 1989)
［Solution］From Vieta's formulas and the given conditions, the two roots of the equation satisfy
$$
\left\{\begin{array}{l}
\operatorname{tg} x_{1}+\operatorname{tg} x_{2}=9 \\
\operatorname{tg} x_{1} \cdot \operatorname{tg} x_{2}=1
\end{array}\right.
$$

Therefore, $\operatorname{tg} x_{1}>0, \operatorname{tg} x_{2}>0$. Also, $x_{1}, x_{2} \in[0,2 \pi]$,
so $x_{1}, x_{2}$ are in the first quadrant, i.e., $0<x_{1}<\frac{\pi}{2}, 0<x_{2}<\frac{\pi}{2}$, and $\quad x_{3}=\pi+x_{1}, \quad x_{4}=\pi+x_{2}$.
Furthermore, $\operatorname{tg}\left(x_{1}+x_{2}\right)=\frac{\operatorname{tg} x_{1}+\operatorname{tg} x_{2}}{1-\operatorname{tg} x_{1} \operatorname{tg} x_{2}}$,
substituting (1) and (2), we find that the denominator is 0. Since $0<x_{1}<\frac{\pi}{2}, 0<x_{2}<\frac{\pi}{2}$, it follows that $x_{1}+x_{2}=\frac{\pi}{2}$, thus $x_{3}+x_{4}=2 \pi+\left(x_{1}+x_{2}\right)=\frac{5 \pi}{2}$, hence
$$
x_{1}+x_{2}+x_{3}+x_{4}=3 \pi \text {. }
$$

Therefore, the answer is $(D)$.