What relationship must hold between the coefficients of two quadratic equations for their roots to form a geometric progression?
Let the two equations be:

$$
a_{1} x^{2}+b_{1} x+c=0, \quad a_{2} x^{2}+b_{2} x+c_{2}=0
$$

If the roots of these equations are $x_{1}, x_{2}, x_{3}$, and $x_{4}$, then they form the following geometric progression:

$$
x_{1}, \quad x_{1} \cdot \frac{x_{2}}{x_{1}}, \quad x_{1} \cdot \frac{x_{2}^{2}}{x_{1}^{2}}, \quad x_{1} \cdot \frac{x_{2}^{3}}{x_{1}^{3}}
$$

Since

$$
x_{3} x_{4}=\frac{c_{2}}{a_{2}}=x_{1}^{2} \cdot \frac{x_{2}^{5}}{x_{1}^{5}}=x_{1} x_{2} \cdot\left(\frac{x_{2}}{x_{1}}\right)^{4}
$$

it follows that

$$
\left(\frac{x_{2}}{x_{1}}\right)^{4}=\frac{a_{1} c_{2}}{a_{2} c_{1}}
$$

Furthermore,

$$
x_{3}+x_{4}=-\frac{b_{2}}{a_{2}}=\frac{x_{1} x_{2}^{2}+x_{2}^{3}}{x_{1}^{2}}=\frac{x_{2}^{2}\left(x_{1}+x_{2}\right)}{x_{1}^{2}}
$$

it follows that

$$
\left(\frac{x_{2}}{x_{1}}\right)^{2}=\frac{a_{1} b_{2}}{a_{2} b_{1}}
$$

Substituting (2) into (1),

$$
\left(\frac{x_{2}}{x_{1}}\right)^{4}=\frac{a_{1}^{2} b_{2}^{2}}{a_{2}^{2} b_{1}^{2}}=\frac{a_{1} c_{2}}{a_{2} c_{1}}
$$

from which

$$
\left(\frac{b_{1}}{b_{2}}\right)^{2}=\frac{c_{1}}{c_{2}} \cdot \frac{a_{1}}{a_{2}}
$$

(Béla Krausz, Pécs.)

The problem was also solved by: Bayer B., Burján K., Czank K., Demeter J., Feldnár V., Filkorn J., Holzmann M., Keesz J., Kerekes T., König D., Krisztián Gy., Kürth A., Lukhaub Gy., Lupsa Gy., Perl Gy., Póka Gy., Rosenberg Á., Scharff J., Schmidl I., Sümegi Gy., Szmodics H., Tézner E.