In this question, \(a\) and \(b\) are distinct, non-zero real numbers, and \(c\) is a real number.
Show that, if \(a\) and \(b\) are either both positive or both negative, then the equation \[\frac{x}{x-a}+\frac{x}{x-b}=1\] has two distinct real solutions.
Show that the equation \[\frac{x}{x-a}+\frac{x}{x-b}=1+c\] has exactly one real solution if \(c^2=-\dfrac{4ab}{(a-b)^2}\). Show that this equation can be written \(c^2=1-\left(\dfrac{a+b}{a-b}\right)^2\) and deduce that it can only hold if \(0 < c^2 \leq 1\).
