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Math Help - Reduce/Expand squareroots and exponentiation

  1. #1
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    Reduce/Expand squareroots and exponentiation

    Hi,

    I've got an to assignment, to reduce/expand the following work:

    (x3 + √12)2 + √4 * 12(1/2) * (x2a + 3 / x2a)

    I've tried to reduce it in pieces.

    (x3 + √12)2 = x6 - 4√3x3
    √4 * 12(1/2) * (x2a + 3 / x2a) = 12 + 4√3x3

    So in all, is it equal to:

    x6 + 4√3x3 + 12 + 4√3x3

    Which is equal to:

    x6+ 8√3x3 + 12

    Though, i'm not sure, that I'm doing it the right way, or if there is more I can do, than the following.

    Thanks in advance.
    Last edited by JrAl; October 9th 2012 at 10:46 AM.
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  2. #2
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    Re: Reduce/Expand squareroots and exponentiation

    Quote Originally Posted by JrAl View Post
    to reduce/expand the following work:

    (x3 + √12)2 + √4 * 12(1/2) * (x2a + 3 / x2a)
    I wish I could read your post. Not sure what is what.

    Here are some guesses.
    (x^3+\sqrt{12})^2=x^6+2\sqrt{12}x^3+12

    \sqrt{4}\sqrt{12}=4\sqrt{3}

    \frac{x^{2a+3}}{x^{2a}}=x^3

    But I don't know what goes with what.
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  3. #3
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    Re: Reduce/Expand squareroots and exponentiation

    Isn't (x3 + √12)2 equal to x6 + 4√3x3 + 12?

    I mean, 4√3x3 instead of 2√3x3.
    Last edited by JrAl; October 9th 2012 at 10:37 AM.
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  4. #4
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    Re: Reduce/Expand squareroots and exponentiation

    Quote Originally Posted by JrAl View Post
    Isn't (x3 + √12)2 equal to x6 + 4√3x3 + 12?

    I mean, 4√3x3 instead of 2√3x3.
    Yes 2\sqrt{12}=4\sqrt{3}
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  5. #5
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    Re: Reduce/Expand squareroots and exponentiation

    All right. Thanks :-)

    So I get:

    x6 + 4√3x3 + 12 + 4√3x3

    Which is equal to x6 + 8√3x3 + 12 ?

    Can I reduce that anymore?
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  6. #6
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    Re: Reduce/Expand squareroots and exponentiation

    Quote Originally Posted by JrAl View Post
    All right. Thanks :-)

    So I get:

    x6 + 4√3x3 + 12 + 4√3x3

    Which is equal to x6 + 8√3x3 + 12 ?

    Can I reduce that anymore?
    NO
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  7. #7
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    Re: Reduce/Expand squareroots and exponentiation

    Thank you for the help!
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