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Math Help - Please simplify and please show the working.........

  1. #1
    varungk92
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    Please simplify and please show the working.........

    square root of 98 + cube root of 32 - sixth root of 50.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by varungk92 View Post
    square root of 98 + cube root of 32 - sixth root of 50.
    \root {2}\of{98} \approx 9.8995

    \root {3}\of{32} \approx 3.1748

    \root {6}\of{50} \approx 1.9194

    so:

    <br />
\root {2}\of{98} + \root {3}\of{32} +\root {6}\of{50} \approx 9.8995+3.1748+1.9194 \approx 14.994<br />

    RonL
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  3. #3
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    Hello, varungk92!

    Are you sure of the wording of the problem?
    It doesn't simplify very much and it's a really stupid problem.

    I suspect that it's suppose to be: . \sqrt{98} + 3\sqrt{32} - 6\sqrt{50}


    You're expected to know how to simplify radicals: . \sqrt{ab} \:=\:\sqrt{a}\cdot\sqrt{b}


    The first term is: . \sqrt{98}\;=\;\sqrt{49\cdot2}\;=\;\sqrt{49}\cdot\s  qrt{2} \;= \;7\sqrt{2}

    The second term is: . 3\sqrt{32}\;=\;3\sqrt{16\cdot2}\;=\;3\sqrt{16}\cdo  t\sqrt{2}\;=\;3\cdot4\cdot\sqrt{2} \;=\;12\sqrt{2}

    The third term is: . 6\sqrt{50}\;=\;6\sqrt{25\cdot2}\;=\;6\cdot\sqrt{25  }\cdot\sqrt{2}\;=\;6\cdot5\cdot\sqrt{2}\;=\;30\sqr  t{2}


    Then the problem becomes: . 7\sqrt{2} + 12\sqrt{2} - 30\sqrt{2}

    Answer: . \boxed{-11\sqrt{2}}

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