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Math Help - How can these roots be real?

  1. #1
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    How can these roots be real?

    I ran this equation

    112u^8+56u^4 v^4+3v^8=48u^8+24u^4 v^4+3v^8

    solving for v

    through http://www.quickmath.com/webMathematica3/quickmath/equations/solve/basic.jsp#v1=3v^8%2B24u^4+v^4%2B48u^8%3D3v^8%2B56u ^4+v^4%2B112u^8&v2=u

    and am confused about it's results. It says that 2 of the values for u are real. Here's one of them (sorry, I couldn't get the indices small with tex so am just using ordinary text

    v=(-1)^(1/4) * 2^(1/4) * u

    I don't understand how (-1)^(1/4) is real. I thought it was the same as \sqrt{i}

    [edit]

    Forgot to add, I've also done this by hand and have 2u^4=-v^4 which I think comes to the same thing.
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  2. #2
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    Re: How can these roots be real?

    Quote Originally Posted by moriman View Post
    I ran this equation

    112u^8+56u^4 v^4+3v^8=48u^8+24u^4 v^4+3v^8

    solving for v

    through http://www.quickmath.com/webMathematica3/quickmath/equations/solve/basic.jsp#v1=3v^8%2B24u^4+v^4%2B48u^8%3D3v^8%2B56u ^4+v^4%2B112u^8&v2=u

    and am confused about it's results. It says that 2 of the values for u are real. Here's one of them (sorry, I couldn't get the indices small with tex so am just using ordinary text

    v=(-1)^(1/4) * 2^(1/4) * u

    I don't understand how (-1)^(1/4) is real. I thought it was the same as \sqrt{i}

    [edit]

    Forgot to add, I've also done this by hand and have 2u^4=-v^4 which I think comes to the same thing.
    The only real solution for u is u=0, in which case v can be anything. The other solutions are all complex, as you correctly say. Moral: Don't always trust free software.

    To get indices to display correctly in TeX, use braces not parentheses (curly brackets rather than round ones, in other words). So for example [TEX]v=(-1)^{1/4} * 2^{1/4} * u[/TEX] yields v=(-1)^{1/4} * 2^{1/4} * u.
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  3. #3
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    Sep 2011
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    Re: How can these roots be real?

    Thanks for the confirmation and for the help on the indices
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