\section*{Problem 4 - 251044}

Determine all non-zero real numbers \(r\) for which the equation

\[
\frac{2 x}{r(x+r)}+\frac{1}{x-2 r}=\frac{4 x-r+6}{r(x-2 r)(x+r)}
\]

a) has exactly two different real solutions, b) has exactly one real solution, c) has no real solution!
}

If \(x \neq -r\) and \(x \neq 2r\), then the equation, when multiplied by the main denominator, is equivalently transformed into

\[4x - r + 6 = (2x)(x - 2r) + r(x + r) = 2x^2 - 4rx + rx + r^2 \quad \text{or} \quad 2x^2 - (3r + 4)x + r^2 + r - 6 = 0\] thus \[x^2 - \frac{3r + 4}{2} \cdot x + \frac{r^2 + r - 6}{2} = 0\]. This quadratic equation in \(x\) has exactly the solutions

\[
\begin{gathered}
x_{1/2} = \frac{3r + 4}{4} \pm \sqrt{\frac{D}{16}} \quad \text{with} \\
D = (3r + 4)^2 - 8(r^2 + r - 6) = 9r^2 + 24r + 16 - 8r^2 - 8r + 48 = r^2 + 16r + 64 = (r + 8)^2 \\
\text{so} \ x_{1/2} = \frac{3r + 4 \pm (r + 8)}{4}, \text{ i.e., } x_1 = \frac{3r + 4 - r - 8}{4} = \frac{r}{2} - 1 \text{ and } x_2 = \frac{3r + 4 + r + 8}{4} = r + 3
\end{gathered}
\]

These two solutions coincide exactly for \(r = -8\).

Now, the apparent solutions of this quadratic equation must be excluded for which \(x = -r\) or \(x = 2r\):

Case 1.1: It is \(-r = x_1 = \frac{r}{2} - 1\). This is equivalent to \(r = 2\).

Case 1.2: It is \(-r = x_2 = r + 3\). This is equivalent to \(r = -\frac{3}{2}\).

Case 2.1: It is \(2r = x_1 = \frac{r}{2} - 1\). This is equivalent to \(r = -\frac{2}{3}\).

Case 2.2: It is \(2r = x_2 = r + 3\). This is equivalent to \(r = 3\).

In summary, the equation in the problem statement has exactly one solution for \(r \in \left\{-8, -\frac{3}{2}, -\frac{2}{3}, 2, 3\right\}\), exactly two solutions for all other non-zero \(r\), and never no solutions.