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Math Help - Simple Indicie problem

  1. #1
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    Simple Indicie problem

    Sorry, very simple question i'm stuck on.

    Example in my book shows:

    5^\frac{1}{2}-5^\frac{3}{2}+5^\frac{5}{2} = \sqrt{5} - 5\sqrt{5} + 5^2\sqrt{5} = (1 - 5 + 25)\sqrt{5} = 21\sqrt{5}

    My Q.

    7^\frac{1}{2}-7^\frac{3}{2}+7^\frac{5}{2} So I just substituted to get  \sqrt{7} - 7\sqrt{7} + 7^2\sqrt{7} = (1 - 7 + 49)\sqrt{7} = 43\sqrt{7}

    Which is way off the answer in my book of 4\sqrt[3]{7}

    thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jordan View Post
    Sorry, very simple question i'm stuck on.

    Example in my book shows:

    5^\frac{1}{2}-5^\frac{3}{2}+5^\frac{5}{2} = \sqrt{5} - 5\sqrt{5} + 5^2\sqrt{5} = (1 - 5 + 25)\sqrt{5} = 21\sqrt{5}

    My Q.

    7^\frac{1}{2}-7^\frac{3}{2}+7^\frac{5}{2} So I just substituted to get  \sqrt{7} - 7\sqrt{7} + 7^2\sqrt{7} = (1 - 7 + 49)\sqrt{7} = 43\sqrt{7}

    Which is way off the answer in my book of 4\sqrt[3]{7}

    thanks!
    43 \sqrt{7} is the right answer (you could have checked this with a calculator). the book made a typo. evidently the small 3 should be a big 3
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  3. #3
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    Hello, Jordan!

    7^\frac{1}{2}-7^\frac{3}{2}+7^\frac{5}{2}

    So I just substituted to get: .  \sqrt{7} - 7\sqrt{7} + 7^2\sqrt{7} \;= \;(1 - 7 + 49)\sqrt{7} \;=\; 43\sqrt{7} . . . . Right!

    Which is way off the answer in my book: . 4\sqrt[3]{7}
    I bet it's a typo in your book.
    . . There is no way that a cube root could appear in that problem!

    Look at your answer and theirs: . 43\sqrt{7}\qquad4\sqrt[3]{7}

    It looks like "The Case of the Misplaced 3" (coming to a theater near you).

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  4. #4
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    Haha, I didn't check it on the calculator because I just assumed the book was right. My mistake.
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