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Math Help - Surds (Q3)

  1. #1
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    Surds (Q3)

    Without using a calculator, find the values of the integers a and b for which the solution of the equation

     x\sqrt{24} = x\sqrt{3} + \sqrt{6}

    is \frac{a+\sqrt{b}}{7}
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  2. #2
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    Quote Originally Posted by Punch View Post
    Without using a calculator, find the values of the integers a and b for which the solution of the equation

     x\sqrt{24} = x\sqrt{3} + \sqrt{6}

    is \frac{a+\sqrt{b}}{7}
    x=\frac{\sqrt{6}}{\sqrt{24}\sqrt{3}}

    Rationalize the surds and you will get

    \frac{\sqrt{144}+\sqrt{18}}{21}

    \frac{12+3\sqrt{2}}{21}<br />
    ...
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  3. #3
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    Hello, Punch!

    \text{Without using a calculator, find the values of }a\text{ and }b

    \text{for which the solution of: }\:x\sqrt{24} \:=\: x\sqrt{3} + \sqrt{6}

    . . \text{is: }\;x\:=\:\dfrac{a+\sqrt{b}}{7}

    We have: . x\sqrt{24} \;=\;x\sqrt{3} + \sqrt{6} \quad\Rightarrow\quad x\sqrt{24} - x\sqrt{3} \;=\;\sqrt{6}

    . . . . . (\sqrt{24}-\sqrt{3})x \;=\;\sqrt{6} \quad\Rightarrow\quad x \;=\;\dfrac{6}{\sqrt{24}-\sqrt{3}}


    Factor and reduce:

    . . x \;=\;\dfrac{\sqrt{2}\!\cdot\!\sqrt{3}}{2\!\cdot\!\  sqrt{2}\!\cdot\!\sqrt{3} - \sqrt{3}} \;=\;\dfrac{\sqrt{3}\!\cdot\!\sqrt{2}}  {\sqrt{3}\!\cdot\!(2\sqrt{2}-1)} \;=\; \dfrac{\sqrt{2}}{2\sqrt{2}-1}


    Rationalize:

    . . x \;=\;\dfrac{\sqrt{2}}{2\sqrt{2}-1}\cdot\dfrac{2\sqrt{2}+1}{2\sqrt{2}+1} \;=\;\dfrac{4 + \sqrt{2}}{8-1} \;=\;\dfrac{4 + \sqrt{2}}{7}


    Therefore: . a = 4,\;\;\rlap{/////}b = 1 \;\;b = 2



    Edit: Right, masters! . . . b = 2 . . . Boy, that was a dumb error!
    Last edited by Soroban; August 20th 2010 at 11:54 AM.
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  4. #4
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    b=2
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