<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] If you multiply both sides by $(x-a)(x-b)$, you get
$$
x(x-b)+x(x-a)=(1+c)(x-a)(x-b),
$$
or $2x^2-(a+b)x=(1+c)(x^2)-(1+c)(a+b)x+(1+c)ab$, or
$(1-c)x^2+c(a+b)x-(1+c)ab=0$. This means that
$$
Ax^2+Bx+C=0,
$$
where
$$
A=1-c,B=c(a+b), C=-(1+c)ab.
$$
This has exactly one real solution iff $B^2-4AC =0\ldots$ You should be able to continue from here.
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