# 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?

The easiest way is to think about it graphically.

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

1. If $$a>0$$, we have $$y>0\iff xr_2$$

2. If $$a<0$$, we have $$y>0\iff r_1

• What about the one root case?
– ABu
Commented Mar 9, 2022 at 15:30
• @Peregring-lk Note there is one unique root when $b^2-4ac=0$. In this case, we have $y=0$ at $x=r$ for a unique $r$, and $y$ is positive elsewhere if $a>0$ while $y$ is negative elsewhere if $a<0$. Commented Mar 9, 2022 at 19:05
• @Peregring-lk The above answer covers the unique root case too; e.g. $\{x:r<x<r\}$ is the empty set. Commented Mar 9, 2022 at 19:15

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.

You can tackle this by completing the square. Notice that \begin{align} 0& 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: $$|a| $$|a|>b \implies a>b \ \ \text{ or } \ a<-b$$ and you're set to isolate $$x$$.

• Shouldn't the second inequality be $x\lt\frac{-b-\sqrt{b^2-4ac}}{2a}$?
– ABu
Commented Mar 8, 2022 at 21:23
• @Peregring-lk Yes, you're right. My apologies. I will fix it. Commented Mar 8, 2022 at 21:26
• @Kman3 You have assumed $a>0$ when diving through by $a$ to obtain $\left(x+\frac{b}{2a}\right)^2>\frac{b^2-4ac}{4a^2}$ Commented Mar 9, 2022 at 5:53
• He never divided through by a, he factored a out. Commented Mar 9, 2022 at 7:44
• @Kman3 After analizing what happens at the $|a|\lt b$ case, can we simplify and say that, for any inequality direction: "the $+\sqrt{}$ side of the solution keeps the original inequality direction, while the $-\sqrt{}$ side inverts it"? If that sentence is true I think you could add it to your answer because it's a rule very easy to remember.
– ABu
Commented Mar 9, 2022 at 7:51