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Math Help - Division with Radicals

  1. #1
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    Division with Radicals

    This is confusing me. The book's method is basically this...
    Problem:
    4
    -
    (Radical) 7

    Multiply by
    (Radical) 7
    -----------
    (Radical) 7

    with (Radical) meaning the checkmark thing, and the hyphens being a division line.
    That multiplication gets you...

    4 * (Radical) 7
    -------------
    7

    Which the book marks as correct.

    That's all well and good, but I'm confused by the next step up...
    What would I do on a problem like this?

    2 * (Radical) 3
    --------------
    (Radical) 6

    I tried that method, and couldn't do it. The book gives no other methods.

    Help please...?
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    Quote Originally Posted by Forkmaster View Post
    That's all well and good, but I'm confused by the next step up...
    What would I do on a problem like this?

    2 * (Radical) 3
    --------------
    (Radical) 6

    I tried that method, and couldn't do it. The book gives no other methods.

    Help please...?
    Hmmm there, you can simplify by \sqrt{3} first, because \sqrt{6}=\sqrt{3} \cdot \sqrt{2}.
    Then, use the given method
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Santa Cruz, CA
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    Quote Originally Posted by Forkmaster View Post
    This is confusing me. The book's method is basically this...
    Problem:
    4
    -
    (Radical) 7

    Multiply by
    (Radical) 7
    -----------
    (Radical) 7

    with (Radical) meaning the checkmark thing, and the hyphens being a division line.
    That multiplication gets you...

    4 * (Radical) 7
    -------------
    7

    Which the book marks as correct.

    That's all well and good, but I'm confused by the next step up...
    What would I do on a problem like this?

    2 * (Radical) 3
    --------------
    (Radical) 6

    I tried that method, and couldn't do it. The book gives no other methods.

    Help please...?
    \frac{2\sqrt{3}}{\sqrt{6}}

    Here is one way of doing this:

    since 2=\sqrt{4} we see that our expression becomes \frac{\sqrt{4}\sqrt{3}}{\sqrt{6}}

    There is a theorem that says \sqrt{a}\cdot\sqrt{b}=\sqrt{ab} given that both a~and~b>0

    So, we then see that our expression becomes \frac{\sqrt{12}}{\sqrt{6}}=\sqrt{\frac{12}{6}}=\co  lor{red}\boxed{\sqrt{2}}

    Here is another:

    Using this idea \sqrt{a}\cdot\sqrt{b}=\sqrt{ab} when a~and~b>0 we can maninpulate the demonimator a little bit:

    \frac{2\sqrt{3}}{\sqrt{6}}=\frac{2\sqrt{3}}{\sqrt{  2\cdot 3}}=\frac{2\sqrt{3}}{\sqrt{2}\sqrt{3}}

    Now, the \sqrt{3} term cancels out.

    So we are left with \frac{2}{\sqrt{2}}

    Now rationalize this expression to get \frac{2}{\sqrt{2}}\cdot{\color{red}\frac{\sqrt{2}}  {\sqrt{2}}}=\frac{2\sqrt{2}}{2}=\color{red}\boxed{  \sqrt{2}}

    I hope this makes sense!

    --Chris

    EDIT: Moo beat me!
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  4. #4
    Newbie
    Joined
    Nov 2007
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    Alright, I get it now. Thanks!

    Edit: Found more that I can't figure out. It's not my day today...>_>

    How would I solve problems like these:

    (Radical) 5
    ----------
    (Radical) 10


    (Radical) 27
    -----------
    (Radical) 45


    (Radical) 32
    -----------
    (Radical) 5


    Or


    (Radical) 18
    -----------
    3 * (Radical) 2
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