What relationship exists between the coefficients of the equation $x^{2}+b x+c=0$ if the square of one root is equal to the negative of the other root: $x_{1}^{2}=-x_{2}$?
I. solution. We can only deal with the question if the roots are real, that is, if $b^{2}-4 c \geq 0$. Based on the requirement, the root denoted by $x_{1}$ can be expressed with $b$ and $c$ using the known relationships between the coefficients and the roots of the quadratic equation, so the sought relationship is given by the requirement that the two expressions must be equal.

Since the coefficient of the quadratic term here is $a=1$, taking into account the desired relationship,

$$
\begin{gathered}
x_{1}+x_{2}=x_{1}-x_{1}^{2}=-b \\
x_{1} x_{2}=-x_{1}^{3}=c
\end{gathered}
$$

From the latter,

$$
x_{1}=\sqrt[3]{-c}
$$

and substituting this into the first, the sought relationship, with appropriate rearrangement,

$$
\sqrt[3]{c}+\sqrt[3]{c^{2}}=b \quad\left(\text { and } b^{2}-4 c \geq 0\right)
$$

In the given case, the fulfillment of the condition can be more conveniently checked in a root-free form. By cubing and again taking (1) into account, we get such a form:

$$
c+3 c\left(\sqrt[3]{c}+\sqrt[3]{c^{2}}\right)+c^{2}=c+3 b c+c^{2}=b^{3}
$$

rearranging,

$$
b^{3}-3 b c-c^{2}-c=0
$$

Remarks. 1. The conditions (1) and (2) are not only necessary (since we started from the fulfillment of the requirement), but also sufficient. Indeed, if we substitute the

$$
b=-\left(x_{1}+x_{2}\right) \quad c=x_{1} x_{2}
$$

relationships into (2), by factoring, we get

$$
x_{1}^{3}+x_{2}^{3}+x_{1}^{2} x_{2}^{2}+x_{1} x_{2}=\left(x_{1}^{2}+x_{2}\right)\left(x_{1}+x_{2}^{2}\right)=0
$$

thus

$$
\text { either } x_{1}^{2}+x_{2}=0, \quad \text { or } x_{2}^{2}+x_{1}=0
$$

so the requirement is satisfied.

Balogh Zoltán (Debrecen, Fazekas M. Gym., II. o. t.)

Füredi Zoltán (Budapest, Móricz Zs. Gym., II. o. t.)

2. The equation demonstrating the desired relationship is of course not written based on (1) or (2), but by choosing $x_{1}$. For example, let $x_{1}=1 / 2$, then $x_{2}=-1 / 4$, and the quadratic equation whose roots are these two numbers is:

$$
(x-1 / 2)(x+1 / 4)=x^{2}-x / 2-1 / 8=0
$$

and the value pair $b=-1 / 2, c=-1 / 8$ indeed satisfies (1).

II. solution. Let $\varepsilon$ be one of the numbers +1 and -1, then

$$
x_{1}=\frac{1}{2}\left(-b-\varepsilon \sqrt{b^{2}-4 c}\right), \quad x_{2}=\frac{1}{2}\left(-b-\varepsilon \sqrt{b^{2}-4 c}\right)
$$

and if the requirement is met, then

$$
\frac{1}{4}\left(-b+\varepsilon \sqrt{b^{2}-4 c}\right)^{2}=\frac{1}{2}\left(b+\varepsilon \sqrt{b^{2}-4 c}\right)
$$

expanding,

$$
b^{2}-b-2 c=\varepsilon(b+1) \sqrt{b^{2}-4 c}
$$

from which, by squaring (where in any case $\varepsilon^{2}=1$), and rearranging, we again arrive at the above (2) condition.

Szekeres Vince (Komárno, Secondary School, II. o. t.)

Remark. The "disappearance" of $\varepsilon$ from the calculation does not mean that (3) is satisfied for any value of $\varepsilon$, but that it is satisfied for some value of $\varepsilon$.