What relationship holds between the given numbers $a, b, c$, if the following equation has a real root

$$
\left(a^{2}+b^{2}+c^{2}\right) x^{2}+2(a-b+c) x+3=0
$$
If a quadratic equation has real roots, then its discriminant is non-negative. (1) The quarter of the discriminant, however, can be transformed into the negative of a sum of squares:

$$
\begin{gathered}
(a-b+c)^{2}-3\left(a^{2}+b^{2}+c^{2}\right)=-2\left(a^{2}+b^{2}+c^{2}+a b-a c+b c\right)= \\
=-\left[(a+b)^{2}+(a-c)^{2}+(b+c)^{2}\right]
\end{gathered}
$$

thus it cannot be positive. Therefore, the value of the discriminant can only be 0, meaning the value of each bracket is 0, that is

$$
a+b=0, \quad a-c=0, \quad b+c=0,
$$

from which the sought relationship $a=c=-b$ follows.

Rácz Éva (Makó, József A. g. I. o. t.)

Remark. The vanishing of the discriminant implies that (1) has only one real root, and according to the found relationship, the equation $a^{2} x^{2}+2 a x+1=(a x+1)^{2}=0$ simplifies to $x=-1 / a$. From the assumption, it follows that the common value of $a, -b$, and $c$ is not 0. (Otherwise, (1) would have no solution.)

Vidovszky István (Budapest, Fazekas M. gyak. g. II. o. t.)