$$
x^{2}-(3 p-2) x+p^{2}-1=0
$$

In the equation above, determine $p$ such that one root is three times the other.
If the roots of the equation are \( x_{1} \) and \( x_{2} \), then

\[
x_{1} + x_{2} = 3p - 2 \text{ and } x_{1} x_{2} = p^{2} - 1
\]

but according to the problem,

\[
x_{1} = 3 x_{2}
\]

so

\[
4 x_{2} = 3p - 2 \text{ and } 3 x_{2}^{2} = p^{2} - 1
\]

from which, eliminating \( x_{2} \), we get:

\[
11 p^{2} - 36 p + 28 = 0
\]

from which

\[
p_{1} = 2, \quad p_{2} = \frac{14}{11}
\]

(Kürti lmre, Eger.)

The problem was also solved by: Ádámffy E., Bánó L., Brámer Á., Chambré M., Dömény I., Égető B., Égető G., Erdős V., Fekete M., Földes R., Friedländer H., Fürchtgott M., Füstös P., Haar A., Harsányi Z., Heimlich P., Jánosy Gy., Kertész G., Kiss J., Koffler B., Kovács Gy., Krampera Gy., Magyari F., Martini J., Messer P., Miklóssy K., Nagy Å., Pám M., Petö L., Ragány B., Rássy P., Riesz M., Rosenberg J., Sárközy P., Schöffer I., Schuster Gy., Schwarz Gy., Sonnenfeld J., Söpkéz Gy., Stagl A., Strobl J., Szécsi I., Szőke D., Tandlich E., Tóth B, the V. district high school math circle.