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Math Help - How to get this other root?

  1. #1
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    How to get this other root?

    Numbers are starting to look fuzzy again...

    So, I have this practice problem:

    x^\frac{1}{2} - 3x^\frac{1}{3} = 3x^\frac{1}{6} - 9

    I was able to get the root x=27

    x^\frac{1}{3}(x^\frac{1}{6} - 3) = 3(x^\frac{1}{6} - 3)
    x^\frac{1}{3} = 3
    x = 27

    However, the book indicates that there are two roots: x = 27 and x = 729. I confirmed this with my calculator, but I am wondering how I was supposed to have figured out that other root. What did I miss here?
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  2. #2
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    Re: How to get this other root?

    Quote Originally Posted by infraRed View Post
    x^\frac{1}{2} - 3x^\frac{1}{3} = 3x^\frac{1}{6} - 9
    I was able to get the root x=27
    x^\frac{1}{3}(x^\frac{1}{6} - 3) = 3(x^\frac{1}{6} - 3)
    x^\frac{1}{3} = 3
    x = 27
    Let y=x^{1/6}
    Can you solve y^3-3y^2=3y-9~?
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: How to get this other root?

    Another possibility is to rewrite:
    x^\frac{1}{3}(x^\frac{1}{6} - 3) = 3(x^\frac{1}{6} - 3)
    as
    x^\frac{1}{3}(x^\frac{1}{6} - 3) - 3(x^\frac{1}{6} - 3)=0
    \Leftrightarrow (x^{\frac{1}{3}}-3)(x^{\frac{1}{6}}-3)=0

    Now use the fact that: a\cdot b=0 \Leftrightarrow a=0 \ \mbox{or} \ b=0
    You can clearly see that there're two solutions.
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