Solve the following equation:

$$
(x-1)(2 x-3)(3 x-4)(6 x-5)=a,
$$

where the value of $a$ is respectively $14 ; 0.1 ; -0.0544$.
Let's factor out the coefficient of $x$ from the two-term factors:

$$
2 \cdot 3 \cdot 6(x-1)\left(x-\frac{3}{2}\right)\left(x-\frac{4}{3}\right)\left(x-\frac{5}{6}\right)=a
$$

and notice that the sum of the first and third factors is equal to the sum of the second and fourth factors:

$$
(x-1)+\left(x-\frac{4}{3}\right)=2 x-\frac{7}{3}=\left(x-\frac{3}{2}\right)+\left(x-\frac{5}{6}\right)
$$

Based on this, we can introduce a new variable as half of this sum:

$$
z=x-\frac{7}{6}
$$

Then, $x=z+7 / 6$, and our equation becomes:

$$
\begin{gathered}
\left(z+\frac{1}{6}\right)\left(z-\frac{1}{3}\right)\left(z-\frac{1}{6}\right)\left(z+\frac{1}{3}\right)=\left(z^{2}-\frac{1}{36}\right)\left(z^{2}-\frac{1}{9}\right)=\frac{a}{36} \\
z^{4}-\frac{5}{36} z^{2}+\frac{1}{324}-\frac{a}{36}=0
\end{gathered}
$$

From this, we get two values for $z^{2}$ - provided the discriminant is positive -

$$
z^{2}=\frac{5 \pm \sqrt{9+144 a}}{72}
$$

From these, we get two roots for $z$ by taking the square root, provided the values of $z^{2}$ are non-negative. Finally, we can calculate $x$ based on the substitution above.

In the first numerical example, $a=14, z^{2}=25 / 36$ and $z^{2}=-5 / 9$. We only get real $z$ from the former: $z= \pm 5 / 6$, and thus $x_{1}=2, x_{2}=1 / 3$.

From $a=0.1$ (to four decimal places), $z^{2} \approx 0.1366$, and $z^{2} \approx 0.0022$, hence $z_{1,2}= \pm 0.370$ and $z_{3,4}= \pm 0.047$, finally $x_{1}=1.536, x_{2}=0.796 ; x_{3}=1.21, x_{4} \approx 1.12$. We get four real roots.

Finally, from $a=-0.0544$, similarly, $z^{2} \approx 0.0844,0.0544, z \approx \pm 0.291, \pm 0.233$, finally $x_{1} \approx 1.458, x_{2} \approx 0.876, x_{3} \approx 1.4$, $x_{4} \approx 0.933=14 / 15$, all four roots are real.

Klara Bornes (Budapest, Blanka Teleki High School, 3rd grade)

Remarks. 1. The key observation in the solution is the same as in the 3rd problem of the 1960 Arany Dániel competition, advanced level, final round. Accordingly, the equation can also be solved using the substitution in the second solution of the competition problem.

2. Let's examine the equation obtained from (1) by multiplication and reduction to 0:

$$
36 x^{4}-168 x^{3}+289 x^{2}-217 x+60-a=0
$$

Based on the result found in problem 656, the polynomial $x^{4}+a x^{3}+b x^{2}+c x+d$ can be written in the form $z^{4}+A z^{2}+B$ if and only if the coefficients satisfy $a^{3}-4 a b+8 c=0$; and if this condition is satisfied, the polynomial can be transformed into the second form by the substitution $x=z-a / 4$. In our case, the coefficients are

$$
-\frac{168}{36}=-\frac{14}{3}, \quad \frac{289}{36}, \quad-\frac{217}{36}
$$

With these, the mentioned condition is satisfied, and the substitution is the one we used above, $x=z+7 / 6$.

The calculation would have been simpler if we had first applied the substitution $x-1=w$. This would have led us to the equation

$$
w(2 w-1)(3 w-1)(6 w+1)-a=36 w^{4}-24 w^{3}-w^{2}+w-a=0
$$

with smaller coefficients. From (2), one can infer that the constant term to be subtracted in each of the four factors is around 1. [^0]

[^0]:    ${ }^{1}$ See the solution in K. M. L. 22 (1961) 5. p.

    ${ }^{2}$ K. M. L. 11 (1955) 57. p.