<s>[INST] solve quadratic inequality for all cases I am in great danger. I am solving a general quadratic inequality $ax^2+bx+c \ge 0$ for $a \neq 0$ as follows:
Let $$ax^2 +bx+c = 0,$$ then the critical points are: $$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}.$$ Then we have three intervals $$\left(- \infty, \frac{-b - \sqrt{b^2-4ac}}{2a}\right], \left[\frac{-b + \sqrt{b^2-4ac}}{2a}, \infty\right), \text{ and } \left[ \frac{-b - \sqrt{b^2-4ac}}{2a},  \frac{-b + \sqrt{b^2-4ac}}{2a}\right].$$
I know that since the inequality is greater than 0, the first two will satisfy the inequality. Hence $$ \left(- \infty, \frac{-b - \sqrt{b^2-4ac}}{2a}\right] \cup \left[\frac{-b + \sqrt{b^2-4ac}}{2a}, \infty\right)$$ is the solution of the given quadratic inequality.
My professor keeps saying "you are wrong, answer me completely and accurately". Anyone, please help me what I am missing and what should be the correct answer?
 [/INST] This answer is nothing more than a consolidation of all of the responses.
To solve: $ax^2 + bx + c \geq 0.$
Since it is presumed that $a \neq 0$, the first thing to do is to break the analysis into two cases: $a < 0$ or $a > 0$.  Then, similar to the manner of deriving the quadratic equation, in each case, you must complete the square.
In both cases, I will use the observation that 
$x^2 + \left(\frac{b}{a}\right)x = \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}.$
I will also use the observation that when $r,s \geq 0$, you have that 
$(r \geq s) \iff (r^2 \geq s^2).$

$\underline{\text{Case 1:} ~a > 0}$
Then $x^2 + \left(\frac{b}{a}\right)x + \frac{c}{a} \geq 0.$
Therefore $\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a} \geq 0.$
Therefore
$$\left(x + \frac{b}{2a}\right)^2 \geq \frac{b^2 - 4ac}{4a^2}.\tag1$$
Within Case 1: there are therefore two possibilities: 
Case 1A:
$b^2 - 4ac \geq 0$.
In this event, based on inequality (1) above, then $x$ must satisfy: 
$\left|x + \frac{b}{2a}\right| \geq \frac{\sqrt{b^2 - 4ac}}{2a}$.
This is satisfied when 
$x \leq - \left(\frac{b}{2a}\right) - \frac{\sqrt{b^2 - 4ac}}{2a}$ 
or 
$x \geq - \left(\frac{b}{2a}\right) + \frac{\sqrt{b^2 - 4ac}}{2a}$.
Case 1B:
$b^2 - 4ac < 0$.
Examining inequality (1) above, since case 1B is presuming that the RHS $< 0$, and since the LHS is always non-negative, inequality (1) will be satisfied for any value of $x$.

$\underline{\text{Case 2:} ~a < 0}$
Then $x^2 + \left(\frac{b}{a}\right)x + \frac{c}{a} \leq 0.$
Therefore $\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a} \leq 0.$
Therefore
$$\left(x + \frac{b}{2a}\right)^2 \leq \frac{b^2 - 4ac}{4a^2}.\tag2$$
Case 2A:
$b^2 - 4ac \geq 0$.
In this event, based on inequality (2) above, then $x$ must satisfy: 
$\left|x + \frac{b}{2a}\right| \leq \frac{\sqrt{b^2 - 4ac}}{|2a|}$.
This is satisfied when 
$x \geq - \left(\frac{b}{2a}\right) - \frac{\sqrt{b^2 - 4ac}}{|2a|}$ 
and 
$x \leq - \left(\frac{b}{2a}\right) + \frac{\sqrt{b^2 - 4ac}}{|2a|}$.
Case 2B:
$b^2 - 4ac < 0$.
Examining inequality (2) above, since case 2B is presuming that the RHS $< 0$, and since the LHS is always non-negative, inequality (2) will never be satisfied for any value of $x$.
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