Posting Freak
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Most of us know the Quadratic Formula and how to use it, but does anyone know where it came from?
I'm trying to Google, but not working well. D:
Posting Freak
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2008-07-06, 09:27 PM
(This post was last modified: 2008-07-06, 09:32 PM by Worthyness.)
From what i remember from my 9th grade summer, I believe you can derive the quadratic formula from ax^2+bx+c. However, i have forgotten that as of now... I am assuming the quadratic equation is :
x=-b+ or - (sqrt(b^2 - 4ac))/ 2a
I'm a bit rusty, it's summer afterall and i haven't used this since the beginning of calculus ><"
Edit:
Did you mean where did it come from as in country/civilization? because i can't help you there...
But if you want to derive the formula, use the completing the square method on the equation
ax^2+bx+c
Sorry if i can't be of any more use =O
Posting Freak
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Let's see if i can do this right...
ax^2+bx+c=0
x^2 + b/a x = -c/a divide by "a"
x^2 + b/a x + ((1/2)(b/a x))^2 = -c/a + ((1/2)(b/a x))^2 completing the square
x^2 + b/a x + b^2/4a^2 = -c/a + b^2/4a^2 simplified
(x^2 + b/2a )^2 = -c/a + b^2/4a^2 Into binomial form (completed square)
x+ b/2a = sqrt(-c/a + b^2/4a^2 ) square root both sides
x= - b/2a + or - sqrt(-c/a + b^2/4a^2 ) move b/a x to other side
x= - b/2a + or - sqrt((-c/a * 4a/4a) + b^2/4a^2 ) common denominator
x= - b/2a + or - (sqrt( b^2 -4ac))/2a simplified
x= (- b + or - sqrt(b^2 - 4ac))/2a simplify more = quadratic formula.
hopefully you can read this... sorry if it is totally incoherent. but i believe that is how it is done... good luck with this =)
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I had an account on AoPS a long time ago...
Does anyone know who first derived that formula?
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I dont know certainly who did it, but I wouldnt doubt it would be the Egyptians. Or Euclid, Pythagoras, or Diophantus.
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that went way over my head ><