Solve the equation

$$
x^{3}-5 x^{2}-9 x+45=0
$$

if we know that two of the roots differ only in sign.
I. solution: The roots of the equation, according to the problem, are $x_{1}, x_{2}=-x_{1}$, and $x_{3}$, and thus its factorized form is

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

By comparing this with the given equation, it follows that

$$
x_{3}=5, \quad x_{1}^{2}=9
$$

and thus

$$
x_{1}=3, \quad x_{2}=-3
$$

Bárdos András (Bp. II., Rákóczi g. III. o. t.)

II. solution: Substitute the two roots with equal absolute values but different signs into the equation

$$
\begin{array}{r}
x_{1}^{3}-5 x_{1}^{2}-9 x_{1}+45=0 \\
-x_{1}^{3}-5 x_{1}^{2}+9 x_{1}+45=0
\end{array}
$$

By adding these two equations,

$$
-10 x_{1}^{2}+90=0
$$

from which

$$
x_{1}^{2}=9 \quad \text { and } \quad x_{1}=3, \quad x_{2}=-3
$$

Dividing the original polynomial equation by the product of the factors $(x-3)$ and $(x+3)$, or $\left(x^{2}-9\right)$,

$$
\left(x^{3}-5 x^{2}-9 x+45\right):\left(x^{2}-9\right)=x-5
$$

Thus, $x-5$ is the third root factor, and so $x_{3}=5$.

Orlik Péter (Bp. V., Eötvös g. I. o. t.)

III. solution: If the equation has rational roots, then these (since the coefficient of $x^{3}$ is 1) can only be integers, and specifically divisors of 45. Since two roots differ only in sign, 45 must also be divisible by the square of these roots. The only square number divisor of 45, apart from 1, is 9. 1 is clearly not a root of our equation, but +3 and -3 are. If two roots are rational, then the third must also be rational, and since $45=3^{2} \cdot 5$, the sought third root is $x_{3}=5$.

Soós Klára (Aszód, Petőfi g. IV. o. t.)

IV. solution: Our equation is naturally solvable without using the simplification given in the problem.

Indeed, our equation can be written as:

$$
x^{2}(x-5)-9(x-5)=0
$$

or

$$
(x-5)\left(x^{2}-9\right)=0
$$

from which either

$$
x^{2}-9=0
$$

or

$$
x-5=0
$$

and thus

$$
x_{1}=3, \quad x_{2}=-3, \quad x_{3}=5
$$

Lábas Elemér (Sátoraljaújhely, Kossuth g. II. o. t.)