In a quadratic

The discriminant is equal to the first derivative.

Does that have any special meaning (geometrically or otherwise), or is it just an algebraic coincidence?

That's not even true...

The first derivative is a function from C->C
The discriminant is a number in C

Don't discriminate against numbers >_<

You should ask this guy I know; he's an expert programmer cause he's read SICP.

AX^2 + BX + C = 0

(+/-)sqrt(B^2 - 4AC) = 2AX + B

manipulates into the quadratic formula

The reason I was asking is because I read it in some vedic math books. Most of the stuff in there becomes common sense after a bit of thinking, but this just seems like it's out there.

(+/-)sqrt(B^2 - 4AC) = 2AX + B

in the above, X is the value which makes the quadratic function zero, whereas in the first derivative, x is a function of y.

also, the discriminant is the expression under the root sign,
d=b^2-4ac

>>6
oops y is a function of x, in other words, x can vary.

mmm, I see what you're saying.  It's probably better phrased as "the discriminant is equal to the derivative at the roots".

What I do know: if the discriminant is 0, then there exists a real, double root.  This also requires the derivative to be 0 at the root.  In this case, they are equal.  This should generalize to higher order polynomials.

What I'm speculating: when the discriminant is not 0, this is probably just an algebraic coincidence.

the quadratic formula is -b/2a (+/-) sqrt(b^2 - 4ac)/ 2a = x

-b/2a is the vertex of the parabola.  The sqrt(b^2 - 4ac) is the distance on each side (Hence +/-) from the vertex to the zero's.

The vertex of a parabola exists when the first derivative is zero.   2ax + b = 0 when X = -b/2a.  Hence the relationship.

Should say sqrt(b^2 - 4ac)/2a not sqrt(b^2 - 4ac)

>>9

Thank you, I see it was all a matter of breaking it down into it's special events and seeing the correlations at those points.

Don't call me gay, but I need some mary jay!

Marijuana MUST be legalized.