## Task 28/73

For which real numbers $a$ and $b$ does the equation

$$
\frac{a}{b-x}-\frac{x}{b+x}=1
$$

have exactly one real solution $x=2$?
If we solve the given equation for $x$, we get

$$
x_{1 ; 2}=\frac{1}{4}\left(b-a \pm \sqrt{a^{2}-10 a b+9 b^{2}}\right.
$$

For there to be exactly one solution, the discriminant must be zero, which implies that either $a=b$ or $a=9b$. For $a=b$, $x=0$, so only $a=9b$ is feasible. From this, we obtain

$$
x=\frac{1}{4}(b-9 b)=-2 b=2
$$

thus $b=1, a=-9$. Therefore, the given equation has exactly the solution $x=2$ for $a=-9, b=1$. A check confirms the correctness.