$$
\begin{gathered}
a x^{2}+b x+c=0 \\
a^{\prime} x^{2}+b^{\prime} x+c^{\prime}=0 \\
\left(a+\lambda a^{\prime}\right) x^{2}+\left(b+\lambda b^{\prime}\right) x+\left(c+\lambda c^{\prime}\right)=0
\end{gathered}
$$

Let the sum of the roots of these equations be denoted by $s, s^{\prime}$, and $S$, and the product of the roots by $p, p^{\prime}$, and $P$. Determine $\lambda$ such that

$$
P=p+p^{\prime}
$$

and express $S$ as a function of $s, s^{\prime}, p$, and $p^{\prime}$.
It is known that

$$
\begin{gathered}
s=x_{1}+x_{2}=-\frac{b}{a}, \quad p=x_{1} x_{2}=\frac{c}{a} \\
s^{\prime}=x_{1}^{\prime}+x_{2}^{\prime}=-\frac{b^{\prime}}{a^{\prime}}, \quad p^{\prime}=x_{1}^{\prime} x_{2}^{\prime}=\frac{c^{\prime}}{a^{\prime}} \\
S=x_{1}^{\prime \prime}+x_{2}^{\prime \prime}=-\frac{b+\lambda b^{\prime}}{a+\lambda a^{\prime}}, \quad P=x_{1}^{\prime \prime} x_{2}^{\prime \prime}=\frac{c+\lambda c^{\prime}}{a+\lambda a^{\prime}}
\end{gathered}
$$

According to the problem,

$$
\frac{c+\lambda c^{\prime}}{a+\lambda a^{\prime}}=\frac{c}{a}+\frac{c^{\prime}}{a^{\prime}}
$$

from which

$$
\lambda=-\frac{a^{2} c^{\prime}}{a^{\prime 2} c}
$$

Substituting this value of \(\lambda\) into the expression for \(S\),

$$
\begin{aligned}
S= & -\frac{b-\frac{a^{2} c^{\prime} b^{\prime}}{a^{\prime 2} c}}{a-\frac{a^{2} c^{\prime} a^{\prime}}{a^{\prime 2} c}}=-\frac{\frac{b}{a}-\frac{a}{c} \cdot \frac{c^{\prime}}{a^{\prime}} \cdot \frac{b^{\prime}}{a^{\prime}}}{1-\frac{a}{c} \cdot \frac{c^{\prime}}{a^{\prime}} \cdot \frac{a^{\prime}}{a^{\prime}}} \\
& =-\frac{-s+\frac{p^{\prime} s^{\prime}}{p}}{1-\frac{p^{\prime}}{p}}=\frac{p s-p^{\prime} s^{\prime}}{p-p^{\prime}}
\end{aligned}
$$

(Zoltán Rehberger, Székesfehérvár.)

The problem was also solved by: D. Barna, I. Bella, F. Bojedain, D. Brandt, M. Devecis, A. Erdős, J. Fekete, E. Freibauer, K. Goldziher, I. Juvancz, J. Kárf, A. Koós, Gy. Krisztián, E. Manheim, R. Obláth, J. Porkoláb, D. Róth, G. Sasvári, E. Schwartz, Ö. Spitzer, I. Szabó, Á. Weisz, J. Weisz.