\section*{Problem 4 - 081224}

All real numbers \(x\) are to be specified for which the equation

\[
|x+1| \cdot|x-2| \cdot|x+3| \cdot|x-4|=|x-1| \cdot|x+2| \cdot|x-3| \cdot|x+4|
\]

is satisfied.
}

For the left side of the equation, we have:

\[
|x+1| \cdot|x-2| \cdot|x+3| \cdot|x-4|= \pm(x+1) \cdot(x-2) \cdot(x+3) \cdot(x-4)
\]

And for the right side:

\[
|x-1| \cdot|x+2| \cdot|x-3| \cdot|x+4|= \pm(x-1) \cdot(x+2) \cdot(x-3) \cdot(x+4)
\]

Thus, from the original equation, we get

Case 1: \((x+1)(x-2)(x+3)(x-4)=(x-1)(x+2)(x-3)(x+4)\) or

Case 2: \((x+1)(x-2)(x+3)(x-4)=-(x-1)(x+2)(x-3)(x+4)\)

We now determine for which (non-integer) values of \(x\) the terms

\[
a=(x+1)(x-2)(x+3)(x-4) \quad \text { and } \quad b=(x-1)(x+2)(x-3)(x+4)
\]

have the same or different signs, i.e., for which values of \(x\) Case 1 or Case 2 should be considered.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1859.jpg?height=349&width=1353&top_left_y=1890&top_left_x=357)

Expanding yields

\[
\begin{aligned}
& (x+1)(x-2)(x+3)(x-4)=x^{4}-2 x^{3}-13 x^{2}+14 x+24 \quad \text { and } \\
& \quad(x-1)(x+2)(x-3)(x+4)=x^{4}+2 x^{3}-13 x^{2}-14 x+24
\end{aligned}
\]

Case 1: To solve is

\[
\begin{gathered}
x^{4}-2 x^{3}-13 x^{2}+14 x+24=x^{4}+2 x^{3}-13 x^{2}-14 x+24 \quad \Longleftrightarrow \quad 4 x^{3}-28 x=0 \\
\Longleftrightarrow \quad x^{3}-7 x=0 \quad \Longleftrightarrow \quad x=0 \text { or } x=\sqrt{7} \text { or } x=-\sqrt{7}
\end{gathered}
\]
\(x_{0}=0\) is clearly a solution to the original equation. For \(x_{1}=\sqrt{7}\), we have \(2\sqrt{\frac{13+\sqrt{64}}{2}}=\sqrt{10.5}>3\). Thus, \(x_{3}\) lies in the Case 2 region and is therefore a solution to the original equation.

Similarly, with \(x_{4}=-\sqrt{\frac{13+\sqrt{73}}{2}}, x_{5}=\sqrt{\frac{13-\sqrt{73}}{2}}\) and \(x_{6}=-\sqrt{\frac{13-\sqrt{73}}{2}}\), we get \(-4<x_{4}<-3, 1<x_{5}<2\) and \(-2<x_{6}<-1\), and hence \(x_{4}, x_{5}\) and \(x_{6}\) are solutions to the original equation.

In total, we have seven solutions \(x_{0}, \ldots, x_{6}\).