<s>[INST] Show that if $c\neq1$, the equation $ \frac{x}{x-a} + \frac{x}{x-b} = 1+c$ has exactly one real solution if $c^2 = - \frac {4ab}{(a-b)^2}$. 
Show that if $c\neq1$, the equation
   $$ \frac{x}{x-a} + \frac{x}{x-b} = 1+c$$ has exactly one real solution if $$c^2 = - \frac {4ab}{(a-b)^2}$$.

I know that a quadratic of the form $ax^2 +bx +c$ has only one real solution if $\sqrt{b^2-4ac} =0$, but I am not sure how to use this to solve the problem that has been set. I have also tried adding the two fractions and rearranging the result, but I have been unable to get anything resembling $ax^2 +bx +c$ .
 [/INST] Since $x \neq a$ and $x \neq b$ you can multiply both sides of the equation by $(x - a)(x - b)$ arriving, after a bit of basic algebra, to
$$(1 - c)x^2 + c(a + b)x - ab(1 + c) = 0$$
Setting the condition $\Delta = 0$ will yield the desired result.
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