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anonymous
 4 years ago
NOT A QUESTION (JUST INTERESTING)
Alternate derivation of the quadratic formula
anonymous
 4 years ago
NOT A QUESTION (JUST INTERESTING) Alternate derivation of the quadratic formula

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You may be familiar with the derivation of the quadratic formula by completing the square...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We can get the exact same result by using the sum and products of roots

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0given a quadratic equation: ax^2+bx+c=0 let the two roots be p and q (which are equal to x)

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1yup, tried that, much interesting ...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What if we found pq? (pq)^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^24ac)/a^2 therefore pq = √(b^24ac) /a

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1did the same way, good,go on.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(p+q)+(pq)=2p=2x (because a root is a solution of x) 2x = b/a + √(b^24ac) /a x = b±√(b^24ac)  2a

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And viola, the quadratic formula

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.0Excellent ! somehow ive never seen this before. thank you !!

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0not satisfied with the completing the square solution huh?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nah, long at messy.. this way seems so much more elegant
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