Condition for $ax^2 + bx + c > 0$ or $ax^2 + bx + c < 0$ In an inequality question I was solving, I got to a step where I was stuck. 
$$\frac{(x-1)(3x-8)}{x^2-3x+4} \ge0$$
I couldn't solve it because I cannot factor the denominator or predict its sign. 
The solution manual says that, $x^2-3x+4>0$ because $b^2-4ac<0$ and $a>0$.
I don't understand this. How do you predict the sign of a quadratic expression from its discriminant and $x^2$ coefficient?
A quick google search didn't proved useful.
 A: If the discriminant is negative, then the polynomial does not have any root, thus its representative curve is always above or always below the $x$-axis, so it has always the same sign which is the sign of its dominating term $ax^2$.
A: If a quadratic equation (in this case $x^2 - 3x + 4 = 0$) has no real solutions, then its value can never change sign as you input different values of $x$. And since, for any ridiculously large values of $x$, the $x^2$ term dominates the other terms (which means that the value of the entire expression has the same sign as just $ax^2$, and therefore the same sign as $a$), the entire expression must have the same sign as $a$ for any value of $x$.
A: You can also tell the sign of that polynomial by "completing the square":
\begin{equation*}
x^2 - 3x + 4 = \left( x - \frac{3}{2} \right)^2 + \frac{7}{4}.
\end{equation*}
Since $x^2 - 3x + 4$ is the sum of a square and a positive number, we have $x^2 - 3x + 4 > 0$ for all $x \in \mathbb{R}$.
A: The Polynomial: 
$ax^2 + bx + c$ has a unique sign if and only if $\Delta= b^2-4ac<0$
Now:
if $a>0$ and  $\Delta= b^2-4ac<0$ then $ax^2 + bx + c>0$
if $a<0$ and  $\Delta= b^2-4ac<0$ then $ax^2 + bx + c<0$
A: The graph of the quadratic function $y=ax^2+bx+c$ is a parabola with hands looking up $(a>0)$ or down $(a<0)$. The parabola may cut the x-axis at two points $(b^2-4ac>0)$, touch the x-axis at one point $(b^2-4ac=0)$ or not touch x-axis $(b^2-4ac<0)$.
The parabola of the quadratic function $y=x^2-3x+4$ has hands looking up $(a>0)$ and does not touch the x-axis $(b^2-4ac>0)$, implying the function's value is always positive.
A: Let p(x) = a$x^2$ + bx + c
=a ($x^2$ +$b\over a$x + $c\over a$) 
=a [ $x^2$ + $b\over a$x + $b^2\over (4a^2)$ - D]
= a ( perfect square - D)

Here D =$$ (discriminant)\over 4a^2$$
