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.

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ax^2+bx+c

ax^2=(-)bx-c

ax^2=(-1)b(x+c/b)

ax^2/b=(-)x-c/b

(ax^2/b)+x=(-)c/b

x(ax/b+1)=(-)c/b

x=(-)(c/b)(ax/b+1)

x=(-)cax/b^2+b)

(-)x^2=ca/(b^2+b)=x^2

x=±sqrt(ca/(b^2+b))

I think that is what you are looking for. Someone double check my math though.

-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)

=====

Another alternative quadratic formula:

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))

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