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anonymous

  • 5 years ago

Can someone help me find the 5 solutions to this equation?

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  1. anonymous
    • 5 years ago
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    \[y ^{5}-100y=0\]

  2. anonymous
    • 5 years ago
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    y=0 is one

  3. anonymous
    • 5 years ago
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    use de moivre's theorem

  4. anonymous
    • 5 years ago
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    I don't know what that is. could you explain please?

  5. anonymous
    • 5 years ago
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    -sqrt(10) 0 sqrt(10) y = -i sqrt(10) y = i sqrt(10)

  6. anonymous
    • 5 years ago
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    Do you know euler's formula?

  7. anonymous
    • 5 years ago
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    where e^(i*pi) = -1

  8. anonymous
    • 5 years ago
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    yeah, I think I learned a bit about that. I don't remember these names though. I don't think my teacher did a good job explaining. thank you

  9. anonymous
    • 5 years ago
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    and e^(ix) = cosx+isin(x)

  10. anonymous
    • 5 years ago
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    to clarify, is this what the answer should look like, or is it in a different format?\[\sqrt[i]{10}\]

  11. anonymous
    • 5 years ago
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    yes

  12. anonymous
    • 5 years ago
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    What class are you in?

  13. anonymous
    • 5 years ago
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    college algebra

  14. anonymous
    • 5 years ago
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    1 or 2 ? because if you are in college algebra, you would use synethic division to solve these things.

  15. anonymous
    • 5 years ago
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    Euler's formula is like pre-calc.. and trig-ish

  16. anonymous
    • 5 years ago
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    1

  17. anonymous
    • 5 years ago
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    I think we are just using substitution to solve for thse. thats all it really says in this chapter of my book

  18. anonymous
    • 5 years ago
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    I think most you need to solve this problem is the ability to factorize the expression, mainly difference of two squares.

  19. anonymous
    • 5 years ago
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    so will i be getting different answers than was given?

  20. anonymous
    • 5 years ago
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    No, the answers given are right! But do you know how to get them?

  21. anonymous
    • 5 years ago
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    not exactly

  22. anonymous
    • 5 years ago
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    Well, first take y as a common factor, you get: \[y(y^4-100)=0\] Now, y^4-100 is a difference of two squares. It can factorized using the formula: a^2-b^2=(a-b)(a+b)

  23. anonymous
    • 5 years ago
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    Ok what you need to do is use factor After you factor to Y (y^2-10) (y^2+10) You can use quadratic formula on the last term (y^2+10) to get the imaginary roots.

  24. anonymous
    • 5 years ago
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    Applying this formula to our equation gives: \[y(y^2-10)(y^2+10)=0\]

  25. anonymous
    • 5 years ago
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    Are you following so far? If something does not make sense to you, let me know before we proceed.

  26. anonymous
    • 5 years ago
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    it looks good so far. Thanks

  27. anonymous
    • 5 years ago
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    Good. Now you have: \[y=0\] \[y^2-10=0 \implies y^2=10 \implies y=\sqrt{10}, y=-\sqrt{10}\] These are three solutions.

  28. anonymous
    • 5 years ago
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    We still have two, which are: \[y^2+10=0 \implies y^2=-10\] What do you think about this last equation?

  29. anonymous
    • 5 years ago
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    hmm. That sort of makes sense.Sorry, i'm not exactly what you mean by what do I think about the last equation?

  30. anonymous
    • 5 years ago
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    OK. You can see that we have y^2=(-)10, while square of a real number is always a positive value. In this case y is a complex number. That's: \[y^2=-10 \implies y=\sqrt{10}i, y=-\sqrt{10}i\] The set of solutions to the given equation is: \[\left\{ 0,\sqrt{10},-\sqrt{10},\sqrt{10}i.-\sqrt{10}i \right\}\]

  31. anonymous
    • 5 years ago
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    It's a comma between the last two numbers, not a point.

  32. anonymous
    • 5 years ago
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    oh ok. Thank you for your help

  33. anonymous
    • 5 years ago
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    You're welcome!

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