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Math Help - radical problems help

  1. #1
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    radical problems help

    can someone check these problem

    1) -3\sqrt[3]{-3} + 2\sqrt[3]{162} + 3\sqrt[3]{81}

    -3\sqrt[3]{-3} = -3\sqrt[3]{3}

    2\sqrt[3]{162} = 6\sqrt[3]{6}

    3\sqrt[3]{81} = 9\sqrt[3]{3}

    = -3\sqrt[3]{3} + 9\sqrt[3]{3} = 6\sqrt[3]{3}

    = 6\sqrt[3]{6} + 6\sqrt[3]{3}


    2) 4\sqrt[6]{3} + 2\sqrt[4]{32} - 3\sqrt[6]{192}
    - 2\sqrt[6]{192}

    4[\sqrt[6]{3} = 4\sqrt[6]{3}

    2\sqrt[4]{32} = 4\sqrt[4]{2}

    3\sqrt[6]{192} = 6\sqrt[6]{3}

    2\sqrt[6]{192} = 4\sqrt[6]{3}

    = 4\sqrt[6]{3} + 4\sqrt[4]{2} - 6\sqrt[6]{3} - 4\sqrt[6]{3}

    = 4\sqrt[4]{2} - 6\sqrt[6]{3}

    3) -\sqrt[3]{320} - 4\sqrt[3]{5} + 2\sqrt[3]{135} + 2\sqrt[3]{16}

    -\sqrt[3]{320} = 4\sqrt[3]{5}

    4\sqrt[3]{5} = 4\sqrt[3]{5}

    2\sqrt[3]{135} = 6\sqrt[3]{5}

    2\sqrt[3]{16} = 4\sqrt[3]{2}

    = 6\sqrt[3]{5} + 4\sqrt[3]{2}
    Last edited by zelda1850; November 14th 2010 at 10:04 AM.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    You have a problem with the images... using LaTeX on here would be easier.

    1) -3\sqrt[6]{3} - 2\sqrt[3]{192} - \sqrt[6]{360}

    -3\sqrt[6]{3} = -3\sqrt[6]{3}

    2\sqrt[3]{192} = 6\sqrt[6]{3}
    This is not right

    \sqrt[6]{360} = 2\sqrt[6]{5}
    This is not right

    3\sqrt[6]{3} + 6\sqrt[6]{3} + 2\sqrt[6]{5}
    This is not right

    3\sqrt[6]{3} - 2\sqrt[6]{5}
    This is not right
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  3. #3
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    my first question i wrote the question wrong and how do i use latex on here
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Use the math tabs.

    Click on Go Advanced and then find the button labelled 'TeX'. Select the text and press the button to see how the tabs are used.
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  5. #5
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    when i clicked tex it shows a math tag around my text but how i write square roots with it
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Hm... to write a 6th root of 192 for example, type in

    \sqrt[6]{192}

    and then wrap the math tags.
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  7. #7
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    e^(i*pi)'s Avatar
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    [tex]\sqrt[n]{f(x)}[/tex] gives \sqrt[n]{f(x)}

    For the more general square root you can go with [tex]\sqrt{f(x)}[/tex] which is \sqrt{f(x)}

    You can also use fractions: [tex]\sqrt{\dfrac{a}{b}}[/tex] gives \sqrt{\dfrac{a}{b}}

    And also brackets [tex]\sqrt[n]{\left(\dfrac{a}{b}\right)^m}[/tex] gives \sqrt[n]{\left(\dfrac{a}{b}\right)^m}
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  8. #8
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    i fixed my first problem can someone check it

    -3 \sqrt[6]{3} - 2 \sqrt[6]{192} - \sqrt[6]{320}

    -3\sqrt[6]{3} = -3\sqrt[6]{3}

    2\sqrt[6]{192} = 2*2\sqrt[6]{3} = 4\sqrt[6]{3}

    \sqrt[6]{320} = 2\sqrt[6]{5}

    = -3\sqrt[6]{3} - 4\sqrt[6]{3} - 2\sqrt[6]{5}

    = -7\sqrt[6]{3} - 2\sqrt[6]{5}
    Last edited by zelda1850; November 14th 2010 at 10:15 AM.
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  9. #9
    MHF Contributor Unknown008's Avatar
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    Where did you get that?

     6\sqrt[6]{192}=   2*3\sqrt[6]{3} = 6\sqrt[6]{3}
    This is not right.

    If that was so, it means \sqrt[6]{192} = \sqrt[6]{3} which is not true.
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  10. #10
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    oh it was a typo suppose to be 2
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  11. #11
    MHF Contributor Unknown008's Avatar
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    Even then,

    2\sqrt[6]{192} \neq 6\sqrt[6]{3}

    2\sqrt[6]{192} = 2\sqrt[6]{3\times 64} = 4\sqrt[6]{3}
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  12. #12
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    oh caculated wrong
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