Let's determine $p$ such that in the equation $x^{2}+p x+2 p=0$, one root is three times the other.

---

Determine $p$ so that in the equation $x^{2}+p x+2 p=0$, one root is three times the other.
The relationships between the roots and coefficients of a quadratic equation are as follows:

$$
x_{1}+x_{2}=x_{1}+3 x_{1}=4 x_{1}=-p
$$

and

$$
x_{1} x_{2}=x_{1} \cdot 3 x_{1}=3 x_{1}^{2}=2 p
$$

From (1), $x_{1}=-\frac{p}{4}$. Substituting this value into (2),

$$
3 \cdot \frac{p^{2}}{16}=2 p
$$

which gives

$$
\begin{gathered}
p_{1}=0 \text { and } p_{2}=\frac{32}{3} \\
\left(\text{The roots of the equation } x^{2}+\frac{32}{3} x+\frac{64}{3}=0 \text { are } x_{1}=-\frac{8}{3}, x_{2}=-8 .\right)
\end{gathered}
$$

Balatoni Ferenc (Bp. VII., Rákóczi közg. k. i. II. o. t.)