Let's calculate the sum $x_{1}^{2}+x_{2}^{2}$, where $x_{1}$ and $x_{2}$ are the roots of the following equation:

$$
x^{2}-(3 a+b) x+2 a^{2}+3 a b-2 b^{2}=0
$$
From the properties of the roots of a quadratic equation, it follows that

$$
x_{1} \cdot x_{2}=2 a^{2}+3 a b-2 b^{2}
$$

$$
x_{1}+x_{2}=3 a+b
$$

If from the square of the second equation we subtract twice the first equation, we get:

$$
x_{1}^{2}+x_{2}^{2}=5\left(a^{2}+b^{2}\right)
$$

(Dévai Erzsébet, Budapest.)

The problem was also solved by: Dávid J., Eckstein V., Fekete M., Friedländer H., Fuchs A., Füstös P., Harsányi Z., Havas E., Heimlich P., Jánosy Gy., Kiss J., Koffler B., Kovács Gy., Lázár B., Meleghy Gy., Miklóssy K., Neumann L., Paunz A., Pichler S., Rajz E., Sárközy E., Schön M., Schuster Gy., Szécsi I., Steiger J., Tandlich E., Tóth B., Vetter E., Wähl V., Wottitz R.