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Math Help - surds

  1. #1
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    surds

    Express \sqrt{66-24\sqrt{6}} in the form of p\sqrt{2}+q\sqrt{3}

    I got p=-3 and q=4 or  p=3 and q=-4

    The answer is  p=-3 and q=4 . My question is why is  p=3 and q=-4 ommitted ??


    Thanks .
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by thereddevils View Post
    Express \sqrt{66-24\sqrt{6}} in the form of p\sqrt{2}+q\sqrt{3}

    I got p=-3 and q=4 or  p=3 and q=-4

    The answer is  p=-3 and q=4 . My question is why is  p=3 and q=-4 ommitted ??


    Thanks .
    That's because \sqrt{66-24\sqrt{6}} is positive (it's a square root.. if it was -\sqrt{66-24\sqrt{6}}, it would surely be negative)

    Now, have a thought on this...
    \sqrt{2}<\sqrt{3}
    Then
    3\sqrt{2}<3\sqrt{3}<4\sqrt{3}

    So if p=3 and q=-4, would p\sqrt{2}+q\sqrt{3} be positive ??
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  3. #3
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    Quote Originally Posted by thereddevils View Post
    Express \sqrt{66-24\sqrt{6}} in the form of p\sqrt{2}+q\sqrt{3}

    I got p=-3 and q=4 or  p=3 and q=-4

    The answer is  p=-3 and q=4 . My question is why is  p=3 and q=-4 ommitted ??


    Thanks .
    Hi

    You effectively get 2 possible solutions since \left(3\sqrt{2}-4\sqrt{3}\right)^2 = \left(-3\sqrt{2}+4\sqrt{3}\right)^2 = 66-24\sqrt{6}

    But \sqrt{66-24\sqrt{6}} > 0 and -3\sqrt{2}+4\sqrt{3} > 0 but 3\sqrt{2}-4\sqrt{3} < 0
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