In the following equation, determine $m$ such that one root is twice the other:

$$
2 x^{2}-(2 m+1) x+m^{2}-9 m+39=0
$$
If one root is $x_{1}$, then the other is $2 x_{1}$; based on the known relationships,

$$
3 x_{1}=\frac{2 m+1}{2} \text { and } 2 x_{1}^{2}=\frac{m^{2}-9 m+39}{2}
$$

Eliminating $x_{1}$ from these two equations, we get

$$
m^{2}-17 m+70=0
$$

from which

$$
m_{1}=10, m_{2}=7
$$

(Vajda Ödön, Budapest.)

The problem was also solved by: Appel S., Barcsay Á., Bayer B., Bender E., Benedek Zs., Burján K., Csete F.A., Czank K., Engel D., Faith F., Fekete N., Filkorn J., Fleischer F., Freibauer E., Jankovich S., Kerekes T., Kiss A., Kohn B., Kornis F., Krausz B., Krisztián Gy., Lindtner M., Lukhaub Gy., Lupsa Gy., Mocsonyi A., Obláth R., Ovenden S., Perl Gy., Pollák L., Rozlosnik P., Sasvári G., Schuller I., Spitzer V., Stern D., Stromfeld F., Szabó J., Szentiványi L., Szibelth S., Sztranyavszky S., Weisz J.