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# how do you derive the alternative quadratic formula from ax^2+bx+c=0??

i would like to know how you derive the alternative quadratic formula from ax^2+bx+c=0.  i know how to derive the quadratic formula but not the alternate.

I would like to know how you derive the alternative quadratic formula from ax^2+bx+c=0. I know how to derive the quadratic formula but not the alternate.

-b ± √(b^2 - 4ac)     b ± √(b^2 - 4ac)
x = ––––––––––––––– * ––––––––––––––
2a                  b ± √(b^2 - 4ac)

-b^2 + b^2 - 4ac
x = –––––––––––––––––
2a(b ± √(b^2 - 4ac))

- 4ac
x = –––––––––––––––––
2a(b ± √(b^2 - 4ac))

- 2c
x = ––––––––––––––
b ± √(b^2 - 4ac)

2c
x = –––––––––––––––
-b ± √(b^2 - 4ac)

=====

f(x) = a(x-h)^2+k
= ax^2-2ahx+ah^2+k
= ax^2+bx+c

So a = a,
-2ah=b, h = -b/2a,
ah^2+k=c, k = c - ah^2

Find zeros:

a(x-h)^2+k = 0

a(x-h)^2 = -k

(x-h)^2 = -k/a

|x-h| = √(-k/a)

x-h = ±√(-k/a)

x = h ± √(-k/a) <== A Quadratic Formula

When you derive the quadratic formula

x1=(-b+sqrt(b^2-4ac))/2a, x2=(-b-sqrt(b^2-4ac))/2a, then to derive the alternative quadratic formula .

x1=(-b+sqrt(b^2-4ac))/2a=(-b+sqrt(b^2-4ac))(-b-sqrt(b^2-4ac))/(2a(-b-sqrt(b^2-4ac)))=2c/(-b-sqrt(b^2-4ac)),

x2=(-b-sqrt(b^2-4ac))/2a=(-b-sqrt(b^2-4ac))(-b+sqrt(b^2-4ac))/(2a(-b+sqrt(b^2-4ac)))=2c/(-b+sqrt(b^2-4ac))