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Thread: Surds (Q3)

  1. #1
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    Surds (Q3)

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

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

    is $\displaystyle \frac{a+\sqrt{b}}{7}$
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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 $\displaystyle a$ and $\displaystyle b$ for which the solution of the equation

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

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

    Rationalize the surds and you will get

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

    $\displaystyle \frac{12+3\sqrt{2}}{21}
    $
    ...
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  3. #3
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    Hello, Punch!

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

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

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

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

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


    Factor and reduce:

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


    Rationalize:

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



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