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Math Help - surds...

  1. #1
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    surds...

    hiya all
    theres this question which i can't seem to get the right answer to...
    3+ square root of 24 / 2 + square root of 6
    you need to put it in the form of a + b square root of c / d

    the answer is 6- square root of 6 / 2 buut i can't get it!!

    thaank you
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  2. #2
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    Quote Originally Posted by katiethegreat View Post
    hiya all
    theres this question which i can't seem to get the right answer to...
    3+ square root of 24 / 2 + square root of 6
    you need to put it in the form of a + b square root of c / d

    the answer is 6- square root of 6 / 2 buut i can't get it!!

    thaank you
    Multiply by it's conjugate which is 2-\sqrt6

    \frac{3+2\sqrt6}{2+\sqrt6} \times \frac{2-\sqrt6}{2-\sqrt6} = \frac{(3+2\sqrt6)(2-\sqrt6)}{(2+\sqrt6)(2-\sqrt6)} = \frac{6-3\sqrt6+4\sqrt6 -12}{4-6}

    I will leave you to simplify that
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  3. #3
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    thank you
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  4. #4
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    Quote Originally Posted by katiethegreat View Post
    hiya all
    theres this question which i can't seem to get the right answer to...
    3+ square root of 24 / 2 + square root of 6
    you need to put it in the form of a + b square root of c / d

    the answer is 6- square root of 6 / 2 buut i can't get it!!

    thaank you
    Hi katiethegreat,

    \frac{3+\sqrt{24}}{2+\sqrt{6}}=\frac{3+\sqrt{6(4)}  }{2+\sqrt{6}}=\frac{3+\sqrt{6}\sqrt{4}}{2+\sqrt{6}  }=\frac{3+2\sqrt{6}}{2+\sqrt{6}}

    To get 2 under the line instead of a surd, multiply the entire fraction by 1,
    in such a way, that you get an integer under the line.
    This means getting rid of the surd part of the denominator.

    To do this, change the sign of one of the 2 values in the denominator to form this fraction.

    \frac{3+2\sqrt{6}}{2+\sqrt{6}}=\frac{3+2\sqrt{6}}{  2+\sqrt{6}}\ \frac{2-\sqrt{6}}{2-\sqrt{6}}=\frac{3+2\sqrt{6}}{2+\sqrt{6}}\ \frac{\sqrt{6}-2}{\sqrt{6}-2}

    Either way leads to the answer.

    Alternatively,

    \frac{3+2\sqrt{6}}{2+\sqrt{6}}=p+q\sqrt{6}

    3+2\sqrt{6}=(2+\sqrt{6})(p+q\sqrt{6})=2(p+q\sqrt{6  })+\sqrt{6}(p+q\sqrt{6})

    =2p+2q\sqrt{6}+p\sqrt{6}+q(6)=(2p+6q)+(p+2q)\sqrt{  6}

    2p+6q=3

    p+2q=2\ \Rightarrow\ 2p+4q=4

    Solving the simultaneous equations gves

    2q=-1\ \Rightarrow\ q=-\frac{1}{2}

    p=3

    \Rightarrow\ \frac{3+2\sqrt{6}}{2+\sqrt{6}}=3-\frac{1}{2}\sqrt{6}=\frac{6-\sqrt{6}}{2}
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