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rvc

  • one year ago

Please Help. I need help with the first two questions. And is the third answer's solution's step is correct?

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  1. rvc
    • one year ago
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  2. hartnn
    • one year ago
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    know the a^3- b^3 formula?

  3. rvc
    • one year ago
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    yep

  4. hartnn
    • one year ago
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    use the property that, in a quadratic equation ax^2+bx+c=0 sum of roots = -b/a product of roots = c/a find the difference of roots from the above two values

  5. hartnn
    • one year ago
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    difference^2 = sum^2 -4*product

  6. anonymous
    • one year ago
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    The third one looks correct. So the solutions would be the appropriate roots of unity. For the second one , write \(x_n\) as \(e^{i\frac{\pi}{2^n}}\) and proceed.

  7. rvc
    • one year ago
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    hmmm.. i gtg now too busy atm ill check these sums later. Thanks to all. If i require help i will surely tag u ppl. :) thanks once again

  8. rvc
    • one year ago
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    for the second one?

  9. rvc
    • one year ago
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    second question? how to do?

  10. anonymous
    • one year ago
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    Like I said, note that \(x_n = e^{i\frac{\pi}{2^{n}}}\) Now multiply the \(x_n\)'s and you will notice an infinite sum in the exponent. Tell me when you get that far.

  11. rvc
    • one year ago
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    @satellite73

  12. Michele_Laino
    • one year ago
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    please, note that, in the first question, we have not a quadratic equation @hartnn

  13. Michele_Laino
    • one year ago
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    @nincompoop please help

  14. Michele_Laino
    • one year ago
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    @dan815 please help

  15. Michele_Laino
    • one year ago
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    @phi please help

  16. Michele_Laino
    • one year ago
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    @Loser66 @Hero please help

  17. Michele_Laino
    • one year ago
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    @mukushla please help

  18. Michele_Laino
    • one year ago
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    what do you think about the first question? @Hero

  19. anonymous
    • one year ago
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    sry Michele what's the problem?

  20. Michele_Laino
    • one year ago
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    if please can help us to solve the first question @mukushla

  21. anonymous
    • one year ago
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    @rvc please re-check the problem 1, it seems incorrect!

  22. dan815
    • one year ago
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    try just plugging in a and b and get expressions for a and b

  23. rvc
    • one year ago
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    the question is the same

  24. dan815
    • one year ago
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    \[a^2-6\sqrt{a}+4=0\\ b^2-6\sqrt{b}+4=0\]

  25. dan815
    • one year ago
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    not sure where to take it from here u can get some expressions for a^3-b^3

  26. dan815
    • one year ago
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    \[a^3=(-4+6\sqrt{a})^{3/2}\\ b^3=(-4+6\sqrt{b})^{3/2}\] or \[(\frac{(a^2+4)}{6})^6=a^3 \\ (\frac{(b^2+4)}{6})^6=b^3\]

  27. dan815
    • one year ago
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    oh i checked ur solution on wolfram i think your question is wrong, because http://www.wolframalpha.com/input/?i=x^2-6*sqrt%28x%29%2B4%3D0 that equation as real roots

  28. Michele_Laino
    • one year ago
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    thanks! for your replies @dan815

  29. dan815
    • one year ago
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    yep sure thing :)

  30. rvc
    • one year ago
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    thank you so much :)

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