Solve the following equation, knowing that two of its roots are reciprocals of each other.

$$
24 x^{3}-10 x^{2}-101 x-60=0 .
$$
I. solution. Let the roots of the equation be $x_{1}, x_{2}, x_{3}$, and the first two be reciprocals of each other, i.e., $x_{1} x_{2}=1$. Then the left side of (1) is identical to the following expression: 1

$$
\begin{gathered}
24\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)= \\
=24 x^{3}-24 x^{2}\left(x_{1}+x_{2}+x_{3}\right)+24 x\left(x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}\right)-24 x_{1} x_{2} x_{3}
\end{gathered}
$$

because the difference between the two is at most a second-degree polynomial, and it vanishes at $x_{1}, x_{2}$, and $x_{3}$, so it can only be identically 0. Therefore, the constant terms are also equal, from which

$$
-24 x_{1} x_{2} x_{3}=-24 x_{3}=-60, \quad \text { hence } \quad x_{3}=5 / 2
$$

Now, from the equality of the terms containing $x^{2}$,

$$
-24 x^{2}\left(x_{1}+x_{2}+x_{3}\right)=-10 x^{2}, \quad x_{1}+x_{2}=\frac{10}{24}-\frac{5}{2}=-\frac{25}{12}
$$

According to these, $x_{1}$ and $x_{2}$ are the roots of the following quadratic equation:

$$
x^{2}-\left(x_{1}+x_{2}\right) x+x_{1} x_{2}=x^{2}+\frac{25}{12} x+1=0
$$

Solving this, the roots of (1) are $x_{1}=-3 / 4, x_{2}=-4 / 3, x_{3}=5 / 2$. The first two are indeed reciprocals of each other.

Nagy László (Esztergom, I. István g. III. o. t.)

Remark. We also arrive at the left side of (4) by dividing the left side of (1) by the product of the factor corresponding to the root obtained from (3) and the coefficient of $x^{3}$, i.e., by $24\left(x-x_{3}\right)=24 x-60$.

II. solution. The factorized form of the left side of (1) is

$$
24\left(x-x_{1}\right)\left(x-\frac{1}{x_{1}}\right)\left(x-x_{3}\right)
$$

and the product of the first two factors is of the form $x+u x+1$, where $u=-x_{1}-1 / x_{1}$. We can determine the value of $u$ from the fact that $x^{2}+u x+1$ is a divisor of the left side of (1), so the remainder of the division is 0 for any value of $x$. Now

$$
\begin{aligned}
& \left(24 x^{3}-\quad 10 x^{2}, \quad-101 x \quad-60\right): \quad\left(x^{2}+u x+1\right)=24 x-(24 u+10)
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_6e63edd687773c8081feg-1.jpg?height=92&width=838&top_left_y=1396&top_left_x=340)

$$
\begin{aligned}
& \frac{\mp(24 u+10) x^{2} \quad \mp\left(24 u^{2}+10 u\right) x \quad \mp(24 u+10)}{\left(24 u^{2}+10 u-125\right) x+(24 u-50)}
\end{aligned}
$$

The remainder vanishes identically if and only if $u$ is such that

$$
24 u^{2}+10 u-125=0 \quad \text { and } \quad 24 u-50=0
$$

According to the latter, only $u=25 / 12$ is possible, and this value also satisfies the first equation.

With this, we again arrive at (4). On the other hand, the quotient of the division, according to (5), is $24\left(x-x_{3}\right)=24 x-24 x_{3}$, so

$$
x_{3}=\frac{-(24 u+10)}{-24}=\frac{50+10}{24}=\frac{5}{2}
$$

III. solution. None of the roots of (1) is 0, because for $x=0$ the value of the left side is -60; thus, the reciprocal of each of the three roots exists. Let us find the equation whose roots are $1 / x_{1}, 1 / x_{2}, 1 / x_{3}$. According to our assumption, for $i=1,2,3$

$$
24 x_{i}^{3}-10 x_{i}^{2}-101 x_{i}-60=0
$$

This can also be written as, since $x_{i} \neq 0$:

$$
-60\left(1 / x_{i}\right)^{3}-101\left(1 / x_{i}\right)^{2}-10\left(1 / x_{i}\right)+24=0
$$

so the equation we are looking for is

$$
-60 x^{3}-101 x^{2}-10 x+24=0
$$

Since according to our notation $1 / x_{1}=x_{2}$ and $1 / x_{2}=x_{1}$, (1) and (6) have two common roots, and the same holds for

$\left.1^{\prime}\right) \quad x^{3}-\frac{10}{24} x^{2}-\frac{101}{24} x-\frac{5}{2}=0 \quad$ and

$$
x^{3}+\frac{101}{60} x^{2}+\frac{1}{6} x-\frac{2}{5}=0
$$

Their factorized forms are

$$
\begin{aligned}
& \left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)=0, \quad \text { and } \\
& \left(x-x_{2}\right)\left(x-x_{1}\right)\left(x-\frac{1}{x_{3}}\right)=0
\end{aligned}
$$

The common roots are the roots of the difference of the two left sides, i.e., the

$$
-\frac{21}{10} x^{2}-\frac{35}{8} x-\frac{21}{10}=-\frac{21}{10}\left(x^{2}+\frac{25}{12} x+1\right)=0
$$

equation, which essentially brings us back to (4). On the other hand, taking the difference of the factorized forms, $\left(1^{\prime \prime}\right)$ and $\left(6^{\prime \prime}\right)$, we get

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

Comparing this with the previous one, we find that $x_{3}$ is a root of the

$$
\frac{1}{x_{3}}-x_{3}=-\frac{21}{10}
$$

equation, i.e., one of the values $5 / 2$ and $2 / 5$. Substitution shows that $x_{3}=5 / 2$ satisfies (1).


[^0]:    ${ }^{1}$ See this topic: Surányi János: Polynomial Identity, K. M. L. 23 (1961/11) 103-105.

    ${ }^{2} \mathrm{The}$ equality of the coefficients of the linear terms also yields the same result.