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Thread: Algebraic number

  1. #1
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    Algebraic number

    Hi.

    End of semester cram time. Professor snuck in the definition of an algebraic number in our assignment.

    I have been able to figure out that

    $\displaystyle 6 + \sqrt[4]{2}$

    is an algebraic number because it is a root of $\displaystyle (x-6)^4-2$

    What about something like:

    $\displaystyle \sqrt{2} + \sqrt{-7}$

    $\displaystyle (x)^2+7 $ and $\displaystyle (x)^2-2$

    Then, my mind goes numb on how to put those together.

    Please help.

    Thank you.
    Last edited by kpizle; Dec 9th 2009 at 06:52 PM. Reason: accidentally submitted when trying to preview
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Let $\displaystyle a = \sqrt{2}+\sqrt{7}$. Then $\displaystyle a^2=2+2\sqrt{14}+7=9+2\sqrt{14}$. Thus $\displaystyle (a^2-9)^2=4\times 14 = 56$ and $\displaystyle a$ is a root of $\displaystyle (x^2-9)^2-56$.
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  3. #3
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    Quote Originally Posted by Bruno J. View Post
    Let $\displaystyle a = \sqrt{2}+\sqrt{7}$. Then $\displaystyle a^2=2+2\sqrt{14}+7=9+2\sqrt{14}$. Thus $\displaystyle (a^2-9)^2=4\times 14 = 56$ and $\displaystyle a$ is a root of $\displaystyle (x^2-9)^2-56$.
    Aha! Exactly what I needed: a systematic approach. I can now do the 15 other ones similar to this.

    Thank you very much!!!!
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    You are very welcome. Good luck!
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