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Math Help - Complex Surds

  1. #1
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    Post [SOLVED] Complex Surds

    Hi everyone. Just found the forum. Revising over surds etc and have hit a snag. Can't figure out how to answer this question:

    You are given that:

    a = \frac{3}{2} , b = \frac{9-\sqrt{17}}{4} and c = \frac{9+\sqrt{17}}{4}

    Show that a + b + c = abc.

    Thanks in advance for any help... I'm tearing my hair out over this one.

    Mo.
    Last edited by j0k3r; September 22nd 2008 at 12:55 PM. Reason: Solved
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  2. #2
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    Therefore:

    \frac{3}{2} + \frac{9-\sqrt{17}}{4} + \frac{9+\sqrt{17}}{4} = \frac{3}{2}*\frac{9-\sqrt{17}}{4}*\frac{9+\sqrt{17}}{4}.


    Mo.
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  3. #3
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    \frac{3}{2} + \frac{9-\sqrt{17}}{4} + \frac{9+\sqrt{17}}{4} = \frac{6*18-17}{4}


    \frac{3}{2} + \frac{9-\sqrt{17}}{4} + \frac{9+\sqrt{17}}{4} = 91


    right? or is it:


    \frac{3}{2} + \frac{9-\sqrt{17}}{4} + \frac{9+\sqrt{17}}{4} = \frac{6*18*-17}{4}


    \frac{3}{2} + \frac{9-\sqrt{17}}{4} + \frac{9+\sqrt{17}}{4} = -1836


    aghhhh >_<
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  4. #4
    Moo
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    Quote Originally Posted by j0k3r View Post
    \frac{3}{2} + \frac{9-\sqrt{17}}{4} + \frac{9+\sqrt{17}}{4} = \frac{6*18-17}{4}
    Hmmm no !

    \frac 32+\frac{9-\sqrt{17}}{4}+\frac{9+\sqrt{17}}{4}=\frac 64+\frac{9-\sqrt{17}}{4}+\frac{9+\sqrt{17}}{4}

    Getting everything on a unique fraction :

    =\frac{6+(9-\sqrt{17})+(9+\sqrt{17})}{4}=\frac{6+18+(\sqrt{17}-\sqrt{17})}{4}=\frac{6+18}{4}

    Agree ?

    By the way, for the RHS : (9-\sqrt{17})(9+\sqrt{17})=9^2-(\sqrt{17})^2=81-17
    thanks to the difference of squares formula : (a-b)(a+b)=a^2-b^2



    I'm looking for another method of solving it, not sure I'll find one >.<
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  5. #5
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    Quote Originally Posted by Moo View Post
    Hmmm no !

    \frac 32+\frac{9-\sqrt{17}}{4}+\frac{9+\sqrt{17}}{4}=\frac 64+\frac{9-\sqrt{17}}{4}+\frac{9+\sqrt{17}}{4}

    Getting everything on a unique fraction :

    =\frac{6+(9-\sqrt{17})+(9+\sqrt{17})}{4}=\frac{6+18+(\sqrt{17}-\sqrt{17})}{4}=\frac{6+18}{4}

    Agree ?

    By the way, for the RHS : (9-\sqrt{17})(9+\sqrt{17})=9^2-(\sqrt{17})^2=81-17
    thanks to the difference of squares formula : (a-b)(a+b)=a^2-b^2



    I'm looking for another method of solving it, not sure I'll find one >.<
    \frac{6}{4} + \frac{9-\sqrt{17}}{4} + \frac{9+\sqrt{17}}{4} = \frac{6}{4}*\frac{9-\sqrt{17}}{4}*\frac{9+\sqrt{17}}{4}.


    I GOT IT Solution for those that want it:

    6+(9-\sqrt{17})+(9+\sqrt{17})=\frac{3}{2}*\frac{9-\sqrt{17}}{4}*\frac{9+\sqrt{17}}{4}*4


    6+18=\frac{3(9-\sqrt{17})}{4*2}*({9+\sqrt{17})}


    8*24=3(81-17)


    192=3(64)


    192=192


    Hell yeh

    Mo.
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  6. #6
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    Hello, j0k3r!

    Are you really having difficulty with Arithmetic?




    Given: . a \:= \:\frac{3}{2},\;\;b \:=\: \frac{9-\sqrt{17}}{4},\;\;c \:= \:\frac{9+\sqrt{17}}{4}

    Show that: . a + b + c \:= \:abc.

    a + b + c \:=\:\frac{3}{2} + \underbrace{\frac{9-\sqrt{17}}{4} + \frac{9 + \sqrt{17}}{4}}

    . . . . . . = \;\frac{3}{2} + \overbrace{\frac{9 - \sqrt{17} + 9 + \sqrt{17}}{4}}

    . . . . . . = \;\frac{3}{2} + \frac{18}{4}

    . . . . . . =\;\frac{3}{2} + \frac{9}{2}

    . . . . . . =\;\frac{12}{2}

    . . . . . . =\;{\color{blue}6}



    a\!\cdot\!b\!\cdot\!c \;=\;\left(\frac{3}{2}\right)\underbrace{\left(\fr  ac{9-\sqrt{17}}{4}\right)\left(\frac{9+\sqrt{17}}{4}\ri  ght)}

    . . . . = \;\left(\frac{3}{2}\right)\overbrace{\left(\frac{8  1 + 9\sqrt{17} - 9\sqrt{17} - 17}{16}\right)}

    . . . . = \;\frac{3}{2}\cdot\frac{64}{16}

    . . . . =\;{\color{blue}6}

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  7. #7
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    Quote Originally Posted by Soroban View Post
    Hello, j0k3r!

    Are you really having difficulty with Arithmetic?
    Uhm no... Did you bother to read the fact that I had solved it? o_O.

    Mo.
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