Here if we consider the above equation to be quadratic then i have got the solution that $a \in [0,8/9)$, but if we consider the above equation to be not quadratic i.e. $a=2$ , then it becomes a linear equation with solution $x = 1/2$ which lies in the above interval.
So I want to ask whether I should include the solution $a=2$ or since the question mentions both the roots so it has to be quadratic? Please guide me.
Draw a graph for $$ f(x)= {2x^2\over (x-1)^2}$$ Now, we need to see for which $a$ both solution of equation $a=f(x)$ are in $(-2,1)$. We
see that $a$ is in $[0,{8\over 9})\cup\{2\}$.
$$(a-2)x^2 - 2ax + a = 0$$
The roots are:
$$r_1,r_2 = \frac{a}{a-2}\mp \frac{ \sqrt{8a}}{2(a-2)}$$
As you've found, to satisfy the quadratic equation, and $-2 <r_1,r_2<1$, we require $0\le a <\frac{8}{9}.$
It may be a matter of semantics, but there is no such thing as "both roots" if there is only one root!
If $a=0$, there is a double root: $r_1=r_2=0.$
