I am facing this problem in common for this type of functions. But let me pick an example to describe that. Say $f:R\rightarrow R,f(x)=x/(1+x^2)$. To find it's range, I am going to find the domain of the inverse function. $$y=x/(1+x^2)$$$$yx^2-x+y=0$$$$x=(1±\sqrt(1-4y^2))/2y...(1)$$ Here $1-4y^2$ must be greater or equal to zero for x being a real output.$$1-4y^2≥0$$$$-1/2≤y≤1/2$$So, this is the domain of my inverse function thus the range of f(x).
But wait a minute, in this range, $y=0$ is also included. But that would make $x=0/0$ in $eq (1)$. which is indetermined. So I can't have $y=0$ in my domain of inverse function thus the range of the main function. But if $f(x) =x/(1+x^2)$ then $f(0)=0$, so there is no reason why I should leave out 0 from my range. And again from the definition of inverse function, we know that the range of the main function would be the domain of the inverse function. So $0$ being present in range f but being absent in the domain of inverse function doesn't seem acceptable.
For every function that includes this type of quadratic equation in the denominator, I find this problem.
The issue is the assumption you make when using the quadratic formula: specifically that it's being applied to a quadratic. Remember, $ax^2 + bx + c$ is a quadratic provided $a \neq 0$. If $a = 0$, the solution is radically different: $x = \frac{-c}{b}$, and nothing else. So, when you get to the step $$yx^2 - x + y = 0,$$ before busting out the quadratic formula, ask first what happens in the case $y = 0$. You'll find that the one solution is $x = 0$. Otherwise, the equation is a quadratic, and you get $$x = \frac{1 \pm \sqrt{1 - 4y^2}}{2y}$$ as before.
If you're careful to deal with edge cases like these, then you won't lose possibilities like $y = 0$.