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Math Help - Difficult algebra..

  1. #1
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    Difficult algebra..

    How can I isolate y? y>0

    55(1-y^20) = (1 - y^10)[30 + 60y^10 + 90y^20]

    I'm having a really tough time. I get to:
    0 = -3y^30 + 2.83 y^20 + y^10 - 0.83

    I have no idea how to solve this kind of equation. Help?
    (I realize I could replace y^10 = x, say, and get a third degree polynomial, but I'm not in the measure of solving such an equation ...)
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  2. #2
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    Quote Originally Posted by Volcanicrain View Post
    How can I isolate y? y>0

    55(1-y^20) = (1 - y^10)[30 + 60y^10 + 90y^20]

    I'm having a really tough time. I get to:
    0 = -3y^30 + 2.83 y^20 + y^10 - 0.83

    I have no idea how to solve this kind of equation. Help?
    (I realize I could replace y^10 = x, say, and get a third degree polynomial, but I'm not in the measure of solving such an equation ...)
    Notice that you can factor the left side as the difference of two squares:

    55(1-y^{20})=55(1-y^{10})(1+y^{10})

    Do you see how you can cancel now?
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  3. #3
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    Wow that's amazing! Thanks a million You really helped me out!
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  4. #4
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    You can now write:

    55(1+y^{10})=30+60y^{10}+90y^{20}

    If you let u=10 you can write this as:


    55(1+y^{u})=30+60y^{u}+90y^{2u}

    The let c=1^u

    So now this takes the form of a quadratic.

    55(1+cy)=30+60cy+90cy^2

    Where c is can be treated as a constant. You should be able to isolate y by factoring or using the quadratic formula.
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  5. #5
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    I was able to, thanks a lot I got the right answer for my problem!!
    You're a godsend!
    Could I directly substitute k = y^10 ?
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  6. #6
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    Quote Originally Posted by Volcanicrain View Post
    I was able to, thanks a lot I got the right answer for my problem!!
    You're a godsend!
    Could I directly substitute k = y^10 ?
    Absolutely
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