where does the quadratic formula come from?

Question

how did they figure out the formula and why does it work?

Answer 1

http://www.purplemath.com/modules/quadform.htm
http://en.wikipedia.org/wiki/Quadratic_formula

Those two sites derive the quadratic formula and explain its history and properties. Anyone else who re-derives the quadratic formula is merely earning 2 points and not answering the question....

Answer 2

You have to imaging that ax^2+bx is the start of a known (x+d)^2 and you have to use the fact that (x-e)(x-e) = x^2-e^2

Divide through by whatever is multiplied on the squared term.
x2 + (b/a)x + c/a
Add and substract what is missing of the known from, the squared term (b^2/4a^2).
x^2 + (b/a)x + (b^2/4a^2) + (c/a) - (b^2/4a^2)
Simplify by converting to a common denominator.
x^2 + (b/a)x + (b^2/4a^2) + (4ac/4a^2) - (b^2/4a^2)
Convert the square form.
(x + b/2a)^2 - (b^2 – 4ac)/4a^2
( x + b/2a + sqrt(b^2 – 4ac)/2a ) ( x + b/2a - sqrt(b^2 – 4ac)/2a )
Here are your two roots, simplified as necessary.
x = [ –b ± sqrt(b^2 – 4ac) ] / 2a

Answer 3

it is derived from the completing the square method.

if you take ax^2 + bx + c = 0, then use the completing the square method, you will see for yourself.

Answer 4

ax^2 + bx + c = 0(a =/= 0)

a(x^2 + bx/a) + c = 0

a(x^2 + bx/a + (b/2a)^2 - (b/2a)^2 ) + c = 0

a(x + b/2a)^2 - b^2/4a + c = 0

a(x+ b/2a)^2 = (4ac - b^2)/4a

(x+ b/2a)^2 = (4ac - b^2)/4a^2

x + b/2a = sqr((4ac - b^2)/4a^2) or -sqr(4ac - b^2)/2a

x + b/2a = sqr(4ac - b^2)/2a or -sqr(4ac - b^2)/2a

x = (-b+sqr(4ac - b^2))/2a or (-b-sqr(4ac - b^2))/2a

Answer 5

ax^2+bx+c=0
multiply both ides by 4a
4a^2x^2+4abx+4c=0
(2ax)^2+2*2x*b+b^2-b^2+4ac=0
(2ax+b)^2=[(b^2-4ac)^2]^1/2
2ax+b=+/-(b^2-4ac)^1/2
2ax=-b+/-(b^2-4ac)^1/2
x=[-b+/-(b^2-4ac)^1/2]/2a