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Math Help - Rearranging

  1. #1
    Newbie SuumEorum's Avatar
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    Rearranging

    Hey everyone, I'm new on this forum and I apologize in advance if I'm posting on the wrong sub-forum .

    I'm trying to rearrange: √18/(√6+√3) to √6(√2-1).

    So I got √18/[√3(√2+1)] which I cancelled down to √6/(√2+1) and then rearranged to √6(√2+1)^-1. Can anyone please tell me what the next step is? Or have I made a mistake already? Thanks.
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    Hello, and welcome !

    Quote Originally Posted by SuumEorum View Post
    Hey everyone, I'm new on this forum and I apologize in advance if I'm posting on the wrong sub-forum .
    Showing your working is already a very good thing. Posting in the wrong forum would be neglectible comparing to it

    I'm trying to rearrange: √18/(√6+√3) to √6(√2-1).

    So I got √18/[√3(√2+1)] which I cancelled down to √6/(√2+1) |||useless ||| and then rearranged to √6(√2+1)^-1. Can anyone please tell me what the next step is? Or have I made a mistake already? Thanks.
    Your working is perfect until the red message, there is the final step missing.


    Remember the identity (a-b)(a+b)=a^2-b^2.
    When you want to "simplify" a denominator containing square roots, this can be very useful because squares of square roots are (in general..) integers!

    So here, try to multiply \frac{\sqrt{6}}{\sqrt{2}+1} above and below (so that it's fair !) by \sqrt{2}{\color{red}-}1


    \frac{\sqrt{6}}{\sqrt{2}+1}=\frac{\sqrt{6}}{\sqrt{  2}+1} \cdot \frac{\sqrt{2}{\color{red}-}1}{\sqrt{2}{\color{red}-}1} \quad \leftarrow \text{that's what I mean}
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  3. #3
    Newbie SuumEorum's Avatar
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    Oh whoah, thanks a lot!
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