can somebody derive the quadratic formula for me?

Answer 1

ax² + bx + c = 0
x² + bx/a + c/a = 0
x² + bx/a = -c/a
x² + bx/a + (b/(2a))² = (b/(2a))²-c/a
(x+b/(2a))² = b²/(4a²)-c/a
(x+b/(2a))² = b²/(4a²)-4ac/(4a²)
(x+b/(2a))² = (b²-4ac)/(4a²)
x+b/(2a) = ±√(b²-4ac)/(2a)
x=(-b±√(b²-4ac))/(2a)

Answer 2

do u want amore interesting way of deriving the formula. I tried this on my own hope u will like it.

ax^2 +bx +c = 0
dividing by a (not =0)
x^2 + (b/a)x +c/a = 0
x*(x+ b/a) + c/a = 0
Substitute [x+(x+b/a)]/2 = u
or x = u-b/2a
So we get
{u-b/2a)}{u-b/2a+b/a} +c/a = 0
{u-b/2a}{u+b/2a} + c/a = 0
u^2 - b^2/(4*a^2) +c/a =0
u^2 = (b^2-4ac)/(4a^2)
u=+or-[sqrt(b^2-4ac)] /2a
x+b/2a = u = +or-[sqrt(b^2-4ac)] /2a
x = -b/2a +or-[sqrt(b^2-4ac)] /2a
x= {-b+or-[sqrt(b^2-4ac)] }/2a

Answer 3

Pascal has deduced it correctly for u and I have nothing more to add

Answer 4

dude, go to this website, it is amazing

http://www.csm.astate.edu/algebra/qform.html

all done