Problem based on Mapping of Functions. 
A function $f:\mathbb{R}\to \mathbb{R}$ is define by, $$f(x) = \frac{\alpha x^2 + 6x - 8}{\alpha + 6x - 8x^2}$$Find the interval of values of $\alpha$ for which $f$ is onto.

So, here is what I did
Let $f(x)$ be $y$
$$y(\alpha + 6x -8x^2) = \alpha x^2 + 6x - 8$$
$$ (\alpha + 8y)x^2 + 6(1-y)x -(8+\alpha y) = 0 $$
Since, $x$ is valid for all Real Numbers,  $ \:36(1-y)^2 + 4(8+ \alpha y)(8y + \alpha) \ge0 $
$$y^2(9+8\alpha) + y(46 + \alpha^2) + (9+8\alpha) \ge0 \tag{1}$$
For $(1)$ to hold for each $y\in\mathbb{R}$, $\:9 + 8\alpha >0 \implies \alpha > -\frac{9}{8}$ 
But that's all I have been able to do and in my book an another condition is given,$\:(46 + \alpha^2)^2 - 4(9 + 8\alpha)^2\le0$. 
I am not able to understand why is the author using this condition, If anyone of you could explain me what the condition has to do with the $\textbf{equation } (1)$, and also if someone could explain me the graphical implication of the condition.
Thank You.
 A: Let me make some comments, that should answer your question:
Firstly, I don't quite understand what you mean by "since $x$ is valid for all Real numbers". It should read, "since the domain of the function $f$ is $\mathbb{R}$".
Nextly, (1) is a quadratic inequality. It says the parabola that represents (1) (I am bit loose here, hope you understand), never crosses y-axis. This is as good as saying, the equation you'll get from (1) has complex roots ( and hence the discriminant is negative, which is the condition that book has given you) and the leading coefficient is positive(which you have done).
Hope this helps.
A: If
$$
ay^2+by+c \ge 0\text{ for all values of }y\in\mathbb{R},
$$
then the discriminant $b^2-4ac$ must be negative or zero, and also $a\ge0$.  If the discriminant is positive, then the quadratic equation $ay^2+by+c=0$ has two real solutions.  The condition $a\ge0$ means the parabola opens upward rather than downward.  So there are two real solutions and the parabola opens upward.  That means that if $y$ is between those two real solutions, then $ay^2+by+c$ is negative.
A: $y$ is in the range of $f$ if and only if your second displayed equation has a real solution. This occurs if and only if equation (1) holds. Equation (1) defines a quadratic equation in the variable $y$ that can only have $0$ as a real root. This happens if and only if its discriminant is less than or equal to 0, which is what the other condition is saying.
(You need the other condition and your condition that $\alpha>-9/8$ to get "$\ge$" in (1))
