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Math Help - Help with surds?

  1. #1
    Junior Member
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    Help with surds?

    Hello,

    Not too sure if this is the correct forum to put this thread into, but I am trying to figure out:

    (- sqrt2 / 3)^2 = 2/9?

    Sorry, its seems basic, but I can't get my head around it?!

    Any help is much appreciated!

    *sqrt is square root!

    Thanks!
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  2. #2
    MHF Contributor
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    West Malaysia
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    Quote Originally Posted by Dranalion View Post
    Hello,

    Not too sure if this is the correct forum to put this thread into, but I am trying to figure out:

    (- sqrt2 / 3)^2 = 2/9?

    Sorry, its seems basic, but I can't get my head around it?!

    Any help is much appreciated!

    *sqrt is square root!

    Thanks!
     <br />
(-\sqrt{\frac{2}{3}})^2=-\sqrt{\frac{2}{3}}\times(-\sqrt{\frac{2}{3}})<br />

    =\frac{2}{3}\times\frac{2}{3}

     <br />
=\frac{2}{9}<br />
    Last edited by mathaddict; August 8th 2009 at 07:23 AM.
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  3. #3
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    Sorry, not too sure if it makes a difference or not, but only the 2 on the top is being square rooted?

    As in:


    (- sqrt2 / 3)^2
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  4. #4
    MHF Contributor
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    Quote Originally Posted by Dranalion View Post
    Sorry, not too sure if it makes a difference or not, but only the 2 on the top is being square rootedAs in:
    (- sqrt2 / 3)^2
    (-1)^2 = -1 * -1 = 1 ; right?

    So (-1sqrt(2))^2 = (sqrt(2))^2 ; right?

    (sqrt(2))^2 = 2 ; 3^2 = 9 : so 2/9 ; kapish?
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