$4 \cdot 20$ There is a $\triangle A B C, a 、 b 、 c$ are the sides opposite to $\angle A 、 \angle B 、 \angle C$ respectively. If $b$ is the arithmetic mean of $a 、 c$, and $\operatorname{tg} \frac{B}{2}$ is the geometric mean of $\operatorname{tg} \frac{A}{2} 、 \operatorname{tg} \frac{C}{2}$, try to form a quadratic equation with $\operatorname{tg} \frac{A}{2}, \operatorname{tg} \frac{C}{2}$ as its roots.
[Solution] From the problem, we have $A+B+C=180^{\circ}$,
$$
2 b=a+c,
$$
$$
\left(\operatorname{tg} \frac{B}{2}\right)^{2}=\operatorname{tg} \frac{A}{2} \cdot \operatorname{tg} \frac{C}{2} .
$$

From (2) and the Law of Sines, $2 \sin B=\sin A+\sin C$,
From (1), $2 \sin (A+C)=\sin A+\sin C$,
which means $2(\sin A \cos C+\cos A \sin C)=\sin A+\sin C$.
Since $\sin x=\frac{2 \operatorname{tg} \frac{x}{2}}{1+\operatorname{tg}^{2} \frac{x}{2}}$,
and $\cos x=\frac{1-\operatorname{tg}^{2} \frac{x}{2}}{1+\operatorname{tg}^{2} \frac{x}{2}},(x=A$ or $C)$,
substituting into (4) and simplifying, we get
$$
\begin{array}{l}
\operatorname{tg} \frac{A}{2}+\operatorname{tg} \frac{C}{2}=3 \operatorname{tg} \frac{A}{2} \operatorname{tg} \frac{C}{2}\left(\operatorname{tg} \frac{A}{2}+\operatorname{tg} \frac{C}{2}\right), \\
\therefore \operatorname{tg} \frac{A}{2} \cdot \operatorname{tg} \frac{C}{2}=\frac{1}{3} .
\end{array}
$$

From (3), $\operatorname{ctg} \frac{A}{2} \cdot \operatorname{ctg} \frac{C}{2}=\left[\operatorname{ctg} \frac{180^{\circ}-(A+C)}{2}\right]^{2}$
$$
=\left(\operatorname{tg} \frac{A+C}{2}\right)^{2} \text {, }
$$

which means $\frac{1}{\operatorname{tg} \frac{A}{2} \cdot \operatorname{tg} \frac{C}{2}}=\left(\frac{\operatorname{tg} \frac{A}{2}+\operatorname{tg} \frac{C}{2}}{1-\operatorname{tg} \frac{A}{2} \cdot \operatorname{tg} \frac{C}{2}}\right)^{2}$.
Substituting (5) into the above equation, we get $\left(\operatorname{tg} \frac{A}{2}+\operatorname{tg} \frac{C}{2}\right)^{2}=\frac{4}{3}$,
since $\operatorname{tg} \frac{A}{2}>0, \operatorname{tg} \frac{C}{2}>0$,
thus $\operatorname{tg} \frac{A}{2}+\operatorname{tg} \frac{C}{2}=\frac{2 \sqrt{3}}{3}$.
By Vieta's formulas, the required equation is
$$
x^{2}-\frac{2 \sqrt{3}}{3} x+\frac{1}{3}=0 \text {, }
$$

which is $3 x^{2}-2 \sqrt{3} x+1=0$.