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Math Help - Powered fractions

  1. #1
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    Powered fractions

    What are the steps to solve this problem?

    x^(2/3) - x^(1/3) - 6 = 0
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  2. #2
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    This is actually a quadratic equation in x^{1/3}.

    Recalling the laws of indices we can say that x^{2/3} = \left(x^{1/3}\right)^2

    To illustrate the point let u = x^{1/3} - thus we have u^2-u-6=0


    Solve the quadratic using your favourite method (this one does factorise) to get two values of u.

    Once you have the values of u you put x^{1/3} = root_1 and x^{1/3} = root_2 where root_1 and root_2 are your values of u. It should then be straightforward to find x
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    Once you have the values of u you put x^{1/3} = root_1 and x^{1/3} = root_2 where root_1 and root_2 are your values of u. It should then be straightforward to find x

    I don't understand that part. I got the two values of u which were -2 and 3.
    Last edited by Joaco; September 12th 2010 at 02:51 PM.
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Joaco View Post
    I don't understand that part. I got the two values of u which were 2 and 3.
    In that case you have u= 2 and u=3

    Since we defined u as being equal to x^{1/3} then x^{1/3} must equal 2 and 3.

    u = x^{1/3} = 2 and u = x^{1/3} = 3.


    You can then solve for x, you will get two solutions but this simply means two values of x satisfy this equation
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  5. #5
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    I'm sorry but how do you solve for x?

    EDIT: Nevermind I wasn't thinking :P

    So I would get x= 8 or x=-27

    Thank you for your help !
    Last edited by Joaco; September 12th 2010 at 02:51 PM.
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