Quadratic inequality with negative roots Assume the following quadratic inequality: $$0\lt x^2+4x-100$$ The solutions are: $$
x\lt -2-\sqrt{104},\qquad x\gt-2+\sqrt{104}
$$
In this case, the positive root keeps the original direction of the inequality ($\gt$), but the negative root inverts it.
However, for the general case: $0\lt ax^2+bx+c$, how do I know which root have which inequality direction? Can I say: "if the root is positive the inequality direction of the solution is the same as the original inequality, or the inverted direction otherwise"? Or does it actually depend on the signs and/or values of $a$, $b$ and $c$? and what about $\geq$? What about quadratic inequations where all of their roots are positive or negative?
 A: The easiest way is to think about it graphically.

If the parabola has real distinct roots $r_1,r_2$ where $r_1<r_2$ (this will be the case if the discriminant is positive, i.e. $b^2-4ac>0$) then

*

*If $a>0$, we have $y>0\iff x<r_1  \text{ or } x>r_2$


*If $a<0$, we have $y>0\iff r_1<x<r_2$
A: Any quadratic inequality can be reduced, after completing the square and some algebra, to one of the forms
$$
(x+r)^2 > s \quad\text{or}\quad (x+r)^2 < s.
$$
If $s<0$, then the first form is true for all $x$ while the second form is true for no $x$. Otherwise, when $s\ge0$, the fact that the square root function is increasing means that those forms are equivalent to
$$
|x+r| > \sqrt s \quad\text{or}\quad |x+r| < \sqrt s
$$
since $\sqrt{z^2}=|z|$. By the definition of the absolute value, the first form resolves to
$$
|x+r| > \sqrt s \quad\text{if and only if }\quad x+r > \sqrt s \text{ or } x+r < -\sqrt s,
$$
while the second form resolves to
$$
|x+r| < \sqrt s \quad\text{if and only if }\quad x+r < \sqrt s \text{ and } x+r > -\sqrt s.
$$
In short, one doesn't have to memorize rules for various types of quadratic inequalities; one can simply use standard algebraic manipulations, carefully.
A: You can tackle this by completing the square. Notice that
$$
\begin{align}
0&<ax^2+bx+c \\ 
&=a\left(x^2+\frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right)+c \\
&=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}+c
\end{align}
$$
which means that
$$a>0 \implies \left(x+\frac{b}{2a}\right)^2>\frac{b^2-4ac}{4a^2} \implies
\left|x+\frac{b}{2a}\right|>\frac{\sqrt{b^2-4ac}}{2a}
$$
$$a<0 \implies \left(x+\frac{b}{2a}\right)^2<\frac{b^2-4ac}{4a^2} \implies
\left|x+\frac{b}{2a}\right|<\frac{\sqrt{b^2-4ac}}{2a}
$$
From there, you simply choose between one of the following inequalities for absolute values:
$$|a|<b \implies -b<a<b$$
$$|a|>b \implies a>b \ \  \text{ or } \ a<-b$$
and you're set to isolate $x$.
