anonymous
  • anonymous
NOT A QUESTION (JUST INTERESTING) Alternate derivation of the quadratic formula
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
You may be familiar with the derivation of the quadratic formula by completing the square...
anonymous
  • anonymous
We can get the exact same result by using the sum and products of roots
anonymous
  • anonymous
yes,,you're right

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anonymous
  • anonymous
given a quadratic equation: ax^2+bx+c=0 let the two roots be p and q (which are equal to x)
hartnn
  • hartnn
yup, tried that, much interesting ...
anonymous
  • anonymous
p+q=-b/c pq=c/a
anonymous
  • anonymous
yes,,that's right
anonymous
  • anonymous
What if we found p-q? (p-q)^2 = p^2 - 2pq + q^2 = (p^2 + q^2) - 2pq = (p+q)^2 - 2pq - 2pq = (p+q)^2 - 4pq = b^2/a^2 - 4*c/a = (b^2-4ac)/a^2 therefore p-q = √(b^2-4ac) /a
anonymous
  • anonymous
sorry, not √ but ±√
hartnn
  • hartnn
did the same way, good,go on.
anonymous
  • anonymous
(p+q)+(p-q)=2p=2x (because a root is a solution of x) 2x = -b/a + √(b^2-4ac) /a x = -b±√(b^2-4ac) ------------- 2a
anonymous
  • anonymous
And viola, the quadratic formula
hartnn
  • hartnn
Good Work !
anonymous
  • anonymous
Thank you
ganeshie8
  • ganeshie8
Excellent ! somehow ive never seen this before. thank you !!
anonymous
  • anonymous
Your welcome
lgbasallote
  • lgbasallote
not satisfied with the completing the square solution huh?
anonymous
  • anonymous
nah, long at messy.. this way seems so much more elegant

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