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nope

-b/a aint even an option...unless you get real lucky lol

it will always be x=frac{-b pm sqrt{b^2-4ac}{2a}
a does not equal zero

that didn't work

lol....its the french version

that only works for integer roots

Then no wonder it's not mathematically sound. Ahaha

well, for roots that dont incvolve radicals

And no, not really. You just gotta tweak it.

so it will equal -b/a, assuming I'm only working with integers?

\[x=\frac\]

no.....-b/a aint even on the board

but i keep trying to find a counterexample and cant

-b/2a is a possibility; but that aint always a root unless the graph just touches once

if its a complete square than -b/2a ia the only root

-b/2a is to find the vertex...

i just want to find a counterexample to -b/a being the sum of the roots

-b/2a is the axis of symmetry; but its also half way between roots :)

my bad, i forgot to mention sum

if its a complete square than -b/2a ia the only root

that was weird lol

3+2=5, forgot to mention that the sum of the roots = -b/a

oh ok

i have a proof

lol, did you just do that?

yes does it make sense?

yea, do you know how it can be useful though?

lol, your welcome XD

I am. My sis just brought this to my attention and I never realized it before. XD

its really cool :)

by way it works for all numbers