Problem source

In this question $a$ and $b$ are distinct, non-zero real numbers, and $c$ is a real number.
\begin{questionparts}
\item Show that, if $a$ and $b$ are either both positive or both negative, then the equation 
\[
\displaystyle \frac {x }{ x-a} + \frac{x }{ x-b} = 1
\]
has two distinct real solutions.
\item Show that, if $c\ne1$, the equation 
\[\displaystyle \frac x { x-a} + \frac{x}{ x-b} = 1 + c\] 
has exactly one real solution if 
$\displaystyle c^2 = - \frac {4ab}{\l a - b \r ^2}$
Show that this condition can be written $\displaystyle c^2= 1 - \l \frac {a+b}{a-b} \r ^2 $ and deduce that it  can only hold if $0 < c^2 \le 1\,$.
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$ \begin{align*}
&& 1 &= \frac{x}{x-a} + \frac{x}{x-b} \\
\Leftrightarrow && (x-a)(x-b) &= x(2x-a-b) \\
\Leftrightarrow && 0 &= x^2-ab
\end{align*}
Therefore if $a,b$ are both positive or both negative, $ab > 0$ and there are two distinct solutions $x = \pm \sqrt{ab}$

\item $\,$ \begin{align*}
&& 1+c &= \frac{x}{x-a} + \frac{x}{x-b} \\
\Leftrightarrow && (1+c)(x-a)(x-b) &= x(2x-a-b) \\
\Leftrightarrow && 0 &= (c-1)x^2-c(a+b)x+ab(1+c) \\
\\
&& 0 &= \Delta = c^2(a+b)^2 - 4 \cdot(c-1)\cdot ab(1+c) \\
&&&= c^2(a+b)^2-4ab(c^2-1) \\
&&&= c^2 ((a+b)^2-4ab)+4ab \\
&&&= c^2(a-b)^2+4ab \\
\Rightarrow && c^2 &= -\frac{4ab}{(a-b)^2} \\
&&&= -\frac{(a+b)^2-(a-b)^2}{(a-b)^2} \\
&&&= 1 - \left ( \frac{a+b}{a-b} \right)^2
\end{align*}

Note that $c^2 \geq 0$ and  $1-x^2 \leq 1$ so $0 \leq c^2 \leq 1$. $c^2 = 0 \Rightarrow ab = 0$, but this is not possible since $a,b \neq 0$, therefore $0 < c^2 \leq 1$
\end{questionparts}