<s>[INST] Discriminant of quadratic formula I noticed something that i find interesting about the quadratic formula. I hope someone can explain it to me.
$f(x) = Ax^2 + Bx + c$
$f'(x) = 2Ax + B$
If I say that $f'(x) = 0$ I get that $x = -\frac{B}{2A}$. From this I get that $f(-\frac{B}{2A})$ is a max/min point. 
The quadratic formula is $x = \frac{-B \pm \sqrt{B^2-4AC}}{2A} = -\frac{B}{2A} \pm \frac{\sqrt{B^2-4AC}}{2A}$. Because $f(-\frac{B}{2A})$ is a max/min point this means that the distance between the max/min point to the points where $f(x)$ is $\frac{\sqrt{B^2-4AC}}{2A}$.
I tried to put$x = -\frac{B}{2A}$ into f(x) and got the following result:$f(-\frac{B}{2A})=A (-\frac{B}{2A})^{2} + B(-\frac{B}{2A}) + C = \frac{AB^2}{4A^2} -
\frac{B^2}{2A} + C = \frac{B^2 -2B^2+4AC}{4A} = \frac{-B^2+4AC}{4A}$
If I take the root of it I get $\frac{\sqrt{4AC-B^2}}{2\sqrt{A}}$. It reminds me of the 2nd part of the quadratic formula$\frac{\sqrt{B^2-4AC}}{2A}$. Are they related? Why?
 [/INST] I follow you up to
$$f\left(-\frac{B}{2A}\right) = \frac{AB^2}{4A^2} - \frac{B^2}{2A} + C = \frac{4AC - B^2}{4A}.$$
But the square root of this gives you a $2\sqrt{A}$ in the denominator, so it's not quite what you have.
Here's the connection I see.  In order to get real roots, $B^2 - 4AC$ needs to be non-negative.
We can rewrite the equation above as
$$f\left(-\frac{B}{2A}\right) = \frac{B^2 - 4AC}{-4A}.$$
We know $A \neq 0$.  If $B^2 - 4AC = 0$, then $f(-B/2A) = 0$, which makes sense: a double real root.
If $B^2 - 4AC > 0$, then the sign of $A$ tells us the sign of $f(-B/2A).$  If $A$ is negative, then $f(-B/2A)$ is positive, and the parabola opens down to intersect the $x$-axis at the two roots.  Likewise, if $A$ is positive, then $f(-B/2A)$ is negative, and the parabola opens up to intersect the $x$-axis at the two roots.
That might not be everything, but that's what I see.
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