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Math Help - Solve Equation w/ Rational Exponent.

  1. #1
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    Solve Equation w/ Rational Exponent.

    I am lost. The Algebra II book I use doesn't explain concepts very well. Does anyone know how to solve this problem as well as clearly explain their chosen steps?

    x^{\frac{1}{2}} -5x^{\frac{1}{4}} +6=0

    Thanks in advance.
    Last edited by Mulya66; January 31st 2007 at 06:14 PM.
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  2. #2
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    Hello, Mulya66!

    I hope there's a typo in the problem.
    . . Otherwise, I can't solve it.
    . . . . . . . . . . . . . . . . . . . . .

    Is it supposed to be: . x^{\frac{2}{3}} -5x^{\frac{1}{3}} + 6 \:=\:0

    Then it factors: . \left(x^{\frac{1}{3}} - 2\right)\left(x^{\frac{1}{3}} - 3\right) \:=\:0

    And we have: . \begin{Bmatrix}x^{\frac{1}{3}} - 2 \:=\:0\quad\Rightarrow\quad x^{\frac{1}{3}}\:=\:2\quad\Rightarrow\quad x\,=\,8 \\<br />
x^{\frac{1}{3}} - 3\:=\:0\quad\Rightarrow\quad x^{\frac{1}{3}}\:=\:3\quad\Rightarrow\quad x\,=\,27 \end{Bmatrix}

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  3. #3
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    Oops! It's all a typo! It's really x^{\frac{1}{2}} -5x^{\frac{1}{4}} + 6 \:=\:0

    (Well, that just shows how tired I am...)

    Quote Originally Posted by Soroban View Post
    Hello, Mulya66!

    I hope there's a typo in the problem.
    . . Otherwise, I can't solve it.
    . . . . . . . . . . . . . . . . . . . . .

    Is it supposed to be: . x^{\frac{2}{3}} -5x^{\frac{1}{4}} + 6 \:=\:0

    Then it factors: . \left(x^{\frac{1}{3}} - 2\right)\left(x^{\frac{1}{3}} - 3\right) \:=\:0

    And we have: . \begin{Bmatrix}x^{\frac{1}{3}} - 2 \:=\:0\quad\Rightarrow\quad x^{\frac{1}{3}}\:=\:2\quad\Rightarrow\quad x\,=\,8 \\<br />
x^{\frac{1}{3}} - 3\:=\:0\quad\Rightarrow\quad x^{\frac{1}{3}}\:=\:3\quad\Rightarrow\quad x\,=\,27 \end{Bmatrix}

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  4. #4
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    Hello, Mulya66!

    x^{\frac{1}{2}} -5x^{\frac{1}{4}} + 6 \:=\:0

    Use the same technique . . .

    It factors: . \left(x^{\frac{1}{4}} - 2\right)\left(x^{\frac{1}{4}} - 3\right) \:=\:0

    Can you finish it now?

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  5. #5
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    Quote Originally Posted by Soroban View Post

    \left(x^{\frac{1}{4}} - 2\right)\left(x^{\frac{1}{4}} - 3\right) \:=\:0
    \begin{Bmatrix}x^{\frac{1}{4}} - 2 \:=\:0\quad\Rightarrow\quad x^{\frac{1}{4}}\:=\:2\quad\Rightarrow\quad x\,=\,16 \\<br />
x^{\frac{1}{4}} - 3\:=\:0\quad\Rightarrow\quad x^{\frac{1}{4}}\:=\:3\quad\Rightarrow\quad x\,=\,81 \end{Bmatrix}

    Thanks! It's very easy now.
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