Problem 1382

I. solution: Let us assume that the positive number $u$ is a root of our equation, that is, the equality

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

holds. We will show that the equation has no root larger than $u$, that is, there is no positive $y$ such that $u+y$ is also a root.

The left side of our equation, with the substitution $x=u+y$ and using (1), can be transformed as follows:

$$\begin{aligned} & y^{3}+(3 u-a) y^{2}+\left(3 u^{2}-2 a u-b\right) y+\left(u^{3}-a u^{2}-b u-c\right)= \\ & \quad=y\left[y^{2}+(3 u-a) y+\left(3 u^{2}-2 a u-b\right)\right]=y\left(y^{2}+p y+q\right) \end{aligned}$$

Here, due to our assumptions, both coefficients are positive, indeed, using (1):

$$\begin{gathered} p=3 u-a=2 u+(u-a)=2 u+\frac{b u+c}{u^{2}}>0 \\ q=3 u^{2}-2 a u-b=u^{2}+u(u-a)+\left(u^{2}-a u-b\right)= \\ =u^{2}+\frac{b u+c}{u}+\frac{c}{u}>0 \end{gathered}$$

therefore, the value of the expression $y\left(y^{2}+p y+q\right)$ is positive for all $y>0$, indeed it is never 0.

Applying our conclusion to the smallest positive root, it follows that it is the only positive root.

A similar transformation can be used to prove the validity of the so-called "Descartes' rule of signs" for any polynomial where there is a single "sign change."

II. solution: Let the roots of the given equation be denoted by $x_{1}, x_{2}, x_{3}$ and expand the polynomial form of the equation in its factored form. Since the coefficient of $x^{3}$ is 1:

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

Therefore, due to the equality of the roots and the first coefficient, all other coefficients of this and the given equation are equal:

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

Let us assume for the moment that all three roots are real and none of them is 0. Then from (3) $c \neq 0$, so the product of the three roots, by our assumption: $c>0$, which is only possible if the number of negative roots is even: 2 or 0, and thus the number of positive roots is 1 or 3. However, the possibility of three positive roots is ruled out by (2), because according to it, the sum of the products of two factors is not positive, here, therefore, the number of positive roots is indeed exactly 1.

If exactly one of the roots, say $x_{3}=0$, then our equations simplify:

$$\begin{gathered} x_{1}+x_{2}=a \\ x_{1} x_{2}=-b(>0) \end{gathered}$$

and from $(2 a)$ it is clear that, as the statement claims, there is one positive and one negative root.

If $x_{2}=x_{3}=0$, then there is nothing to prove.

If, however, our equation has a complex root, say $x_{1}=p+q i$, where $p$ and $q$ are real and $q \neq 0$, that is, the following holds:

$$\left(p^{3}-a p^{2}-b p-c-3 p q^{2}+a q^{2}\right)+\left(3 p^{3}-2 a p-b-q^{2}\right) q i=0$$

which means that both bracketed expressions are 0 (we assume, of course, that $a, b, c$ are real), then the conjugate of $x_{1}$, $x_{2}=p-q i$, which is not equal to $x_{1}$, is also a root, so at most one of the three roots can be positive.

Judit Nagy (Szombathely, Kanizsai Dorottya Secondary School, 3rd grade)

Remark: Some "second solutions" write the equation in the form $x^{3}=a x^{2}+b x+c$ and try to understand the number of positive roots based on the graphs of the curves $y=x^{3}$ and $y=a x^{2}+b x+c$ as follows.

"The real roots are given by the abscissas of the common points of the two curves. The curve $y=x^{3}$ rises from the third quadrant through the origin into the first quadrant, and we are only interested in the arc in the first quadrant. For positive $x$, the curve $y=a x^{2}+b x+c$ also rises everywhere (more precisely: it does not sink anywhere); indeed, if $a \neq 0$, then the transformation

$$y=a\left(x+\frac{b}{2 a}\right)^{2}+\frac{4 a c-b^{2}}{4 a}$$

indicates that the vertex of the parabola is "below" and at the non-positive $x_{0}=-\frac{b}{2 a}$ abscissa; if $a=0$ and $b \neq 0$, then $b>0$ and the line $y=b x+c$ rises for all $x$; finally, if $a=b=0$, the line $y=c(\geq 0)$ is parallel to the $X$-axis. Furthermore, since the curve $y=a x^{2}+b x+c$ enters the first quadrant at the point on the $Y$-axis with ordinate $c \geq 0$, the two curves can have only one common point in the first quadrant."

!

In this reasoning, only the last sentence is incorrect; it is not certain that two rising curve arcs can intersect at most once, as they can have any number of common points by "bending" as shown in our diagram. The authors of such solutions implicitly thought of the smoothness of the curves, their tendency to bend little. The first solution essentially showed that there can be no further common point to the right of the common point with abscissa $u$, because for all $x>u$, the ordinate of the curve $y=x^{3}$ is greater than the ordinate of the curve $y=a x^{2}+b x+c$.

Let us note as a lesson: intuition often provides good ideas, but it usually applies to specific cases and can lead to hasty generalizations.