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  1. ali2x2
    • one year ago
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    i saw tat question but i couldt help ;/

  2. ali2x2
    • one year ago
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    thats cheap asto

  3. ganeshie8
    • one year ago
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    there is really a very simple and cute solution, give it a try again :)

  4. ali2x2
    • one year ago
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    ok :)

  5. ParthKohli
    • one year ago
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    inb4 Astrophysics tags me

  6. ParthKohli
    • one year ago
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    These are perfect squares. The only way we have it is that each one is zero.

  7. ganeshie8
    • one year ago
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    Thats it!

  8. ali2x2
    • one year ago
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    I couldnt find a way but now that i see the correct answer it clears up the fog in my mind, good job @ParthKohli :)

  9. ganeshie8
    • one year ago
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    please finish it off @ParthKohli i think @praxer is still looking for a solution

  10. ParthKohli
    • one year ago
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    Oh, sorry. OS is acting up for me again. I type out things and it removes them.

  11. anonymous
    • one year ago
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    this suddenly look like ultimate troll question after you explain it lol

  12. ganeshie8
    • one year ago
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    Haha if you don't like the simplicity, there is a complicated solution, which is quite enlightening too :) Familiar with Cauchy-Schwarz inequality ?

  13. ParthKohli
    • one year ago
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    IT REMOVED IT AGAIN! Yes, I actually solved the same question on my test.

  14. anonymous
    • one year ago
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    no im not at that stage of mathematics yet

  15. Astrophysics
    • one year ago
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    \[\sum_{i=1}^{n} (a_ix+b_i)^2 = x^2 \sum_{i=1}^{n}a^2_i+2x \sum_{i=1}^{n}a_ib_i+\sum_{i=1}^{n}b_i^2=0\]

  16. ParthKohli
    • one year ago
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    The OP isn't actually too far from directly solving it. You can just expand and complete the square.

  17. ganeshie8
    • one year ago
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    what do you mean by directly expand and complete the square ? i thought we will have to use cauchy-schwarz inequality

  18. Astrophysics
    • one year ago
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    Yeah same, can you show it parth

  19. Astrophysics
    • one year ago
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    Oh I think you mean what op was already doing

  20. ganeshie8
    • one year ago
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    Ohkiee, I'll finish off your solution using geometry : |dw:1438963298665:dw|

  21. ganeshie8
    • one year ago
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    \[\sum_{i=1}^{n} (a_ix+b_i)^2 = x^2 \sum_{i=1}^{n}a^2_i+2x \sum_{i=1}^{n}a_ib_i+\sum_{i=1}^{n}b_i^2=0\] then discriminant is given by : \[\left(2\sum_{i=1}^{n}a_ib_i\right)^2-4\left( \sum_{i=1}^{n}a^2_i\right)\left( \sum_{i=1}^{n}b^2_i\right) \]

  22. ganeshie8
    • one year ago
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    thats what OP had too, so we're on same page :)

  23. Astrophysics
    • one year ago
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    perfect

  24. Astrophysics
    • one year ago
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    \[\le 0\]

  25. ganeshie8
    • one year ago
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    Ahh right, but whats the justification for \(\le 0\) ?

  26. Astrophysics
    • one year ago
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    Well I think since \[\sum_{i=1}^{n} (a_ix+b_i)^2 = 0\] is only way we can get real roots, so each term in the sum will have to be 0, and I don't really know the exact proof of it goes, but if we want what you have above \[\left(2\sum_{i=1}^{n}a_ib_i\right)^2-4\left( \sum_{i=1}^{n}a^2_i\right)\left( \sum_{i=1}^{n}b^2_i\right)\] this would need to be <= 0, for the roots of \[x^2 \sum_{i=1}^{n}a^2_i+2x \sum_{i=1}^{n}a_ib_i+\sum_{i=1}^{n}b_i^2=0\] exist

  27. ganeshie8
    • one year ago
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    That looks good! since the quadratic \(\sum_{i=1}^{n} (a_ix+b_i)^2 \) is nonnegative, its determinant must be \(\le 0\) : \[\left(2\sum_{i=1}^{n}a_ib_i\right)^2-4\left( \sum_{i=1}^{n}a^2_i\right)\left( \sum_{i=1}^{n}b^2_i\right) \le 0\] We just proved Cauchy-Schwarz inequality!

  28. Astrophysics
    • one year ago
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    Haha, xD nice!

  29. ali2x2
    • one year ago
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    Why is this still up and viewing xD

  30. ganeshie8
    • one year ago
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    one of the cool things in math in general is TMTOWTDI ;p

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