Intuitive meaning behind the Discriminant While I was applying the quadratic formula, it just suddenly dawned to me that the quadratic formula was basically finding the vertex and adding or subtracting (the plusminus) half the distance between the roots. $\frac{-b}{2a}$ represents the vertex and the $\pm \frac{\sqrt{\triangle}}{2a}$ represents half the distance between the roots. I understand why $-\frac{b}{2a}$ is the xcoor of the vertex, it intuitively makes sense. But why does the $\frac{\sqrt{\triangle}}{2a}$ equal half the distance between the roots? i know how to prove it but I dont why it intuitively works. 
Edit: I understand why the $\frac{\sqrt{\triangle}}{2a}$  is half the distance from the vertex but why is the discriminant B^2 - 4ac. I know thats what happens when you complete the square but when you use algebra you often lose the intuitive sense of what's happening. 
 A: It may be helpful to think about this in terms of symmetries and transformations of graphs.  Let $f(x) = x^2$.  The graph of $f$ is our primitive parabola.  Observer that $f$ is an even function, i.e.
$$ f(-x) = x^2 = f(x) $$
for all real numbers $x$.  This means that the graph of $f$ is symmetric about the $y$-axis.  The vertex of this graph is at the origin.
Now consider shifting this graph down $k$ units.  That is, consider the graph of the function $x \mapsto x^2 - k$, where $k \ge 0$.  As this is just a basic transformation, it doesn't affect the underlying symmetry of the graph.  Indeed, this new function is still symmetric across the $y$-axis.  But observe that
$$ x^2 - k = (x-\sqrt{k})(x+\sqrt{k}), $$
hence the roots of this function are $\pm\sqrt{k}$.  Also observe that, thanks to the underlying symmetry, the two roots are equidistant from the vertex, which implies that the midpoint of the segment joining the roots is on the axis of symmetry.
Next, shift this graph horizontally.  For some real number $h$, consider the function $x \mapsto (x-h)^2 + k$.  This moves the vertex to the point $(h,k)$, but preserves the basic symmetry of the graph—the axis of symmetry is now the line $x = h$, and the roots are $h\pm k$.  As before, the midpoint of the roots is on the axis of symmetry.
Finally, a completely general quadratic function is given by
$$ A(x-h)^2 + k.$$
This is obtained by scaling the previous function by a factor of $A$ (and absorbing a factor of $A$ into the vertical shift).  We still haven't done anything that breaks the original symmetry, hence the roots of this (which are given by
$$ x = h \pm \sqrt{-\frac{k}{A}}, $$
from which we can obtain the quadratic formula, if we like) are again equidistant from the vertex, and have midpoint on the axis of symmetry. 
A: That is because, when you complete the square, $\;\Bigl(x+\dfrac b{2a}\Bigr)^2$ is the square of the $x$-distance from the point with abscissa $x$ to the point with abscissa $-\dfrac b{2a}$, i.e. to the vertex of the parabola.
A: When you complete the square you can rewrite the equation $ax^2+bx+c=0$ (with $a>0$) as
$$
\left(x+\frac{b}{2a}\right)^{\!2}=\frac{b^2-4ac}{4a^2}
$$
If you set $X=x+\frac{b}{2a}$ then the equation gets the form
$$
X^2=\frac{b^2-4ac}{4a^2}
$$
For an equation $X^2=c$, the difference of the roots (larger minus smaller) is obviously $2\sqrt{c}$; a shift of the unknown doesn't change the difference of the roots. So the difference of the roots is
$$
2\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\sqrt{b^2-4ac}}{a}
$$
If you only assume $a\ne0$, the difference should be written
$$
\frac{\sqrt{b^2-4ac}}{|a|}
$$
A: The discriminant also has a fine kinematic interpretation. Note that the coefficients are the parameters of an initial value problem for the function, namely: $f(0)=c, f'(0)=b, f''(x)=2a.$ Asking if $f$ has any roots (when the problem is not trivial) is the same as asking if a particle with initial velocity $v=b$ and constantly decelerated by $g=2a$ would travel a distance of at least $c$ before turning back. (I am assuming, for the sake of clarity, that $a$ and $b$ have opposite signs; otherwise you just have to reverse the direction of time.) The distance traveled is given by Torricelli's equation:
$$ \Delta (v^2) = 2g \Delta s$$
$$-b^2 = 4a \Delta s$$
Have you ever noticed the similarity between this formula and the discriminant?
Now, $b^2 - 4ac = -4a(\Delta s + c)$, so the discriminant is positive if and only if the distance from the $x$-axis at the vertex $(k = \Delta s + c)$ has the opposite sign as that of $a$ (meaning the particle accelerates from the vertex towards the $x$-axis). I think this makes intuitive the relationship between the discriminant and the question of existence and number of roots.
But what is it? If $f$ means position, then the discriminant has units of speed squared. What is the meaning of this speed? According to Torricelli's equation, $4ac$ equals the change in the velocity squared of the particle as it makes a displacement $c$; while $-b^2$ equals the change in the velocity squared of the particle from $x=0$ to the turning point or vertex. So the discriminant is how much more change in the velocity squared that the particle actually undergoes on its way to reach the vertex, than it would have to undergo to travel the distance to the $x$-axis. No wonder it tells you if there are roots!
$\sqrt{\Delta}$ can be understood as the extra change in velocity (beyond needed to reach the axis). I think this makes it more or less intuitive that this is precisely equal to the change in velocity between a root and the vertex (if there is a root). And then, if you divide it by acceleration you get that $\frac{\sqrt{\Delta}}{2|a|}$ is half of the distance between the roots.
Unfortunately for educational purposes, this will not satisfy students who have not learned Calculus: “That doesn't explain anything! Deducing Torricelli's equation means precisely finding the vertex of a quadratic function, so you'd have to do the discriminant there anyway!” Although if you have learned Calculus, you know that this is as simple as $(v^2)'=2vg$. At the very least, you will have an intuitive image to go along with the expression of the determinant.
