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Math Help - Fractional indices

  1. #1
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    Fractional indices

    Help needed with the following question

    x^{\frac{1}{3}} + 9x^{-\frac{1}{3}} = 4

    Solve for x .
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  2. #2
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    Multiply through by x^{1/3}

    (x^{1/3})^2 + 9 = 4x^{1/3} and rearrange to give \displaystyle \left(x^{1/3}\right)^2 - 4x^{1/3} + 9 = 0 which is a quadratic equation.
    Last edited by e^(i*pi); December 17th 2010 at 05:37 AM. Reason: latex
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  3. #3
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    Please help in solving that quadratic....I' getting stuck.

    I got

     Let  y=x^{\frac{1}{3}}<br /> <br />
          Then,<br />
          y^{2}-4y+9=0

    I dont know what to do after that..... I used the quadratic formula...but that didnt work....it's giving imaginary values but the answer consists of real values
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  4. #4
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    Quote Originally Posted by Arka View Post
    Please help in solving that quadratic....I' getting stuck.

    I got

     Let  y=x^{\frac{1}{3}}<br /> <br />
          Then,<br />
          y^{2}-4y+9=0

    I dont know what to do after that..... I used the quadratic formula...but that didnt work....it's giving imaginary values but the answer consists of real values
    Everything has been done correctly - you must have typed the question wrong.
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  5. #5
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    Quote Originally Posted by Arka View Post
    Please help in solving that quadratic....I' getting stuck.

    I got

     Let  y=x^{\frac{1}{3}}<br /> <br />
          Then,<br />
          y^{2}-4y+9=0

    I dont know what to do after that..... I used the quadratic formula...but that didnt work....it's giving imaginary values but the answer consists of real values
    Yes, that is correct- completing the square, y^2- 4y+ 4- 4+ 9= 0 becomes (y- 2)^2= -5. That has NO real roots which means that the original equation, x^{1/3}- 9x^{-1/3}= 4has no real roots. Either you have copied the equation incorrectly or the answer is "there is no real number solution".
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