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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.
 

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Colin L. | Film and Video Production Tutor (Math+Science+Reading too!)Film and Video Production Tutor (Math+Sc...
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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.
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
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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

Yajing L. | Ph.D student from Colorado State UniversityPh.D student from Colorado State Univers...
0
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))