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Math Help - Rationalise

  1. #1
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    Rationalise

    Rationalise

    Could someone show most of the main steps in achieving the rationalised result of
    I tried many methods and failed.
    thank you
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  2. #2
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    e^(i*pi)'s Avatar
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    Did you multiply top and bottom by the conjugate of the denominator?

    \dfrac{(2-\sqrt{2})-\sqrt{3}}{(2+\sqrt{2})-\sqrt{3}} \cdot \dfrac{(2+\sqrt{2})+\sqrt{3}}{(2+\sqrt{2})+\sqrt{3  }} = \dfrac{(4-2)+ (2-\sqrt{2})\sqrt{3} - (2+\sqrt{2})\sqrt{3}-3}{4 + 2\sqrt{2}+2-3}

    = \dfrac{2\sqrt{3}-\sqrt{6}-2\sqrt{3}+\sqrt{6}-1}{3+2\sqrt{2}} = -\dfrac{1}{3+2\sqrt{2}}


    You should be able to multiply by \dfrac{3-2\sqrt{2}}{3-2\sqrt{2}} to rationalise
    Last edited by e^(i*pi); February 3rd 2011 at 04:39 AM. Reason: make latex shorter and thus clearer
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