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Math Help - Radicals #7

  1. #1
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    Post Radicals #7

    Question #3 a)b)c)d): imgur: the simple image sharer

    My attempt: imgur: the simple image sharer

    I have no clue how I should start c) or d), I tried a) and b) but got it wrong...
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  2. #2
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    Re: Radicals #7

    Hello, Mathnood768!

    Simplify. Identify the variable for which the radical represents a real number.

    3(a)\;\;\text{-}3\left(\sqrt{2} - 4\right) + 9\sqrt{2}
    We have: . \text{-}3\sqrt{2} + 12 + 9\sqrt{2} \;=\;12 + 6\sqrt{2}



    (b)\;\;7\left(\text{-}1 - 2\sqrt{6}\right) + 5\sqrt{6} + 8
    We have: . -7 - 14\sqrt{6} + 5\sqrt{6} + 8 \;=\;1 - 9\sqrt{6}



    (c)\;\;4\sqrt{5}\left(\sqrt{3j} + 8\right) - 3\sqrt{15j} + \sqrt{5}
    We have: . 4\sqrt{15j} + 32\sqrt{5} - 3\sqrt{15j} + \sqrt{5} \;=\;\sqrt{15j} + 33\sqrt{5}

    If \sqrt{15j} is real, then j \ge 0




    (d)\;\;3 - \sqrt[3]{4k}\left(12 + 2\sqrt[3]{8}\right)

    We have: . 3 - \sqrt[3]{4k}(12 + 2\!\cdot\!2) \;=\;3 - \sqrt[3]{4k}(12 + 4) \;=\;3 - 16\sqrt[3]{4k}

    If \sqrt[3]{4k} is real, then k = any real number.
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  3. #3
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    Re: Radicals #7

    Thank you!
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