Find the complete set of values of a such that (x^2-x)/(1-ax) all rea…

$y=\frac{x^2-x}{1-ax}$

or $x^2- x=y-axy$

or $x^2+x (ay-1)-y=0$

Since x is real we get

$(ay - 1)^2+ 4y ≥0$

or $a^2y^2+2y(2-a) +1≥0\, ∀\,y∈ R$

So, as $a^2 > 0$,

$4(2-a)^2-4a^2≤0$

or $4-4a≤0$

or $a∈ [1,∞)$