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Math Help - what is the next step?

  1. #1
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    what is the next step?

    square root of 6 (11 square root of 2 minus 2 square root of 2)

    I did everything in the paranthesis first

    (11 square root of 2 minus 2 square root of 2)

    (9 square root of 2)

    (which breaks down to 3 square root of 2)

    so far I got square root of 6(3 square root of 2)

    what do I do next?
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  2. #2
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    1.  \sqrt{6} \left( 11\sqrt{2} - 2\sqrt{2} \right)
    \sqrt{6} \left( 9\sqrt{2} \right)
    9\sqrt{6 \cdot 2} = 9\sqrt{3 \cdot 2 \cdot 2} = 18\sqrt{3}

    How did it break down to 3\sqrt{2} ???
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  3. #3
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    Quote Originally Posted by Chop Suey View Post
    1.  \sqrt{6} \left( 11\sqrt{2} - 2\sqrt{2} \right)
    \sqrt{6} \left( 9\sqrt{2} \right)
    9\sqrt{6 \cdot 2} = 9\sqrt{3 \cdot 2 \cdot 2} = 18\sqrt{3}

    How did it break down to 3\sqrt{2} ???
    How did you get 9 square root of 3 times 2 times 2??????
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  4. #4
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    \sqrt[n]{ab} = \sqrt[n]{a} \cdot \sqrt[n]{b}

    Example:
    \sqrt{2} \cdot \sqrt{6} = \sqrt{6 \cdot 2} = \sqrt{12}
    \sqrt{3} \cdot \sqrt{9} = \sqrt{3 \cdot 9} = \sqrt{27}

    But:

    \sqrt[3]{3} \cdot \sqrt{2} \neq \sqrt{2 \cdot 3}\ \text{or}\  \sqrt[3]{2 \cdot 3}
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  5. #5
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Laura901 View Post
    square root of 6 (11 square root of 2 minus 2 square root of 2)

    I did everything in the paranthesis first

    (11 square root of 2 minus 2 square root of 2)

    (9 square root of 2)

    (which breaks down to 3 square root of 2)


    so far I got square root of 6(3 square root of 2)

    what do I do next?
    take note that 9 is not inside the square root.. so, you should not take the root of 9..
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