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 $ax^2 +bx +c$ .

A first attempt might be to write $\frac{x}{x-a}=\frac{x-a}{x-a}+\frac{a}{x-a}=1+\frac{a}{x-a}$ and so the equation is equivalent to $\frac{a}{x-a}+\frac{b}{x-b}=c-1$.
Now, you can multiply both sides with $(x-a)(x-b)$ and simplify to get the usual quadratic form.
