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

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2 Answers

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)

=====

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))