For what value of the parameter $p$ will the sum of the squares of the roots of the equation

$$
x^{2}+(3 p-2) x-7 p-1=0
$$

be the smallest? What is this smallest value?
I. Solution: If the roots are $x_{1}, x_{2}$, then using the relationship between the roots and coefficients of a quadratic equation:

$$
\begin{gathered}
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(3 p-2)^{2}-2(-7 p-1)= \\
=9 p^{2}+2 p+6=\left(3 p+\frac{1}{3}\right)^{2}+\frac{53}{9}
\end{gathered}
$$

The value on the right side is non-negative, and it is minimal when

$$
3 p+\frac{1}{3}=0 ; \quad \text { i.e., } \quad p=-\frac{1}{9}
$$

Since at this value of $p$ the first term is 0, the smallest value of the sum of squares is

$$
\frac{53}{9}=5 \frac{8}{9}
$$

Domokos Szász (Bp. V., Eötvös g. II. o. t.)

II. Solution: The sum of the squares of the roots can also be calculated differently. The two roots of the equation clearly satisfy the equation, so:

$$
x_{1}^{2}+(3 p-2) x_{1}-7 p-1=0
$$

and

$$
x_{2}^{2}+(3 p-2) x_{2}-7 p-1=0
$$

Add the two equations and note that

$$
\begin{gathered}
x_{1}+x_{2}=-(3 p-2): \\
x_{1}^{2}+x_{2}^{2}=(3 p-2)^{2}+2(7 p+1)= \\
=9 p^{2}+2 p+6=\left(3 p+\frac{1}{3}\right)^{2}+\frac{53}{9}
\end{gathered}
$$

The rest of the solution proceeds identically to the first solution.

Gábor Szatmári (Bp. VIII., Piarista g. II. o. t.)