<s>[INST] Find values of $a$ such that $\frac{x^2-x}{1-ax}$ attains all real values 
Find values of $a$ such that
  $$\frac{x^2-x}{1-ax}$$
  attains all real values.

I first tried to find the range of the above by equating the above to some $y$. Solving I get $x^2+x(ay-1)-y=0$. How do I proceed?
 [/INST] Suppose $f(x)=\frac{x^2-x}{1-ax}=k$ for some fixed constants $a,k$. Solving for $x$ we get
$$x^2-x=k(1-ax)=k-kax$$
$$x^2+(ka-1)x-k=0$$
$$x=\frac{1-ka\pm\sqrt{(ka-1)^2+4k}}2$$
For a chosen $a$, $f$ attains all real numbers if and only if the discriminant $(ka-1)^2+4k$ is non-negative for all $k$. Expanding:
$$(ak)^2-2ka+1+4k\ge0$$
$$a^2k^2+(4-2a)k+1\ge0$$
Using the discriminant a second time we find
$$(4-2a)^2-4(a^2)(1)\le0$$
$$16-16a+4a^2-4a^2\le0$$
$$16\le16a$$
$$a\ge1$$
If $a=1$, $f(x)$ behaves like $-x$. However, it has a removable singularity at $x=1$, which means it does not attain a value of $-1$. Therefore $a=1$ is excluded, and $\frac{x^2-x}{1-ax}$ attains all real values when $a>1$.
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