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apple_pi Group Title

NOT A QUESTION (JUST INTERESTING) Alternate derivation of the quadratic formula

  • one year ago
  • one year ago

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  1. apple_pi Group Title
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    You may be familiar with the derivation of the quadratic formula by completing the square...

    • one year ago
  2. apple_pi Group Title
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    We can get the exact same result by using the sum and products of roots

    • one year ago
  3. cornitodisc Group Title
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    yes,,you're right

    • one year ago
  4. apple_pi Group Title
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    given a quadratic equation: ax^2+bx+c=0 let the two roots be p and q (which are equal to x)

    • one year ago
  5. hartnn Group Title
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    yup, tried that, much interesting ...

    • one year ago
  6. apple_pi Group Title
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    p+q=-b/c pq=c/a

    • one year ago
  7. cornitodisc Group Title
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    yes,,that's right

    • one year ago
  8. apple_pi Group Title
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    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

    • one year ago
  9. apple_pi Group Title
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    sorry, not √ but ±√

    • one year ago
  10. hartnn Group Title
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    did the same way, good,go on.

    • one year ago
  11. apple_pi Group Title
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    (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

    • one year ago
  12. apple_pi Group Title
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    And viola, the quadratic formula

    • one year ago
  13. hartnn Group Title
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    Good Work !

    • one year ago
  14. apple_pi Group Title
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    Thank you

    • one year ago
  15. ganeshie8 Group Title
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    Excellent ! somehow ive never seen this before. thank you !!

    • one year ago
  16. apple_pi Group Title
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    Your welcome

    • one year ago
  17. lgbasallote Group Title
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    not satisfied with the completing the square solution huh?

    • one year ago
  18. apple_pi Group Title
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    nah, long at messy.. this way seems so much more elegant

    • one year ago
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