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Math Help - Express sum of fractional indices in the form k(sqrt 2)

  1. #1
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    Express sum of fractional indices in the form k(sqrt 2)

    I can't solve the following question:

    Simplify 2^\frac{-3}{2}+2^\frac{3}{2}+2^\frac{-1}{2}+2^\frac{1}{2}, giving the answer in the form k\sqrt{2}.

    I tried solving for k, but got nowhere, and I can't think of anything else to do.

    Am I missing something dreadfully obvious here?
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  2. #2
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    Wink A thought

    you already know that 2^{-3/2} equals 1/2^{3/2} and 2^{-1/2} equals 1/2^{1/2}. After changing them over, group them together and group the other two terms with each other and then do some factoring which will get you over there.
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  3. #3
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    Quote Originally Posted by EvilKitty View Post
    I can't solve the following question:

    Simplify 2^\frac{-3}{2}+2^\frac{3}{2}+2^\frac{-1}{2}+2^\frac{1}{2}, giving the answer in the form k\sqrt{2}.

    I tried solving for k, but got nowhere, and I can't think of anything else to do.

    Am I missing something dreadfully obvious here?
    Rewrite the sum as sum of square-roots:

    2^\frac{-3}{2}+2^\frac{3}{2}+2^\frac{-1}{2}+2^\frac{1}{2} = \sqrt{\frac18}+\sqrt{8}+\sqrt{\frac12}+\sqrt{2}

    Now calculate the roots partially (if possible):

    \frac14 \sqrt{\frac12}+2\sqrt{2}+\frac12\sqrt{2}+\sqrt{2} = \frac18 \sqrt{2}+2\sqrt{2}+\frac12\sqrt{2}+\sqrt{2}

    Can you take it from here?
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  4. #4
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    Hello, EvilKitty!

    Simplify: 2^{-\frac{3}{2}}+2^{\frac{3}{2}}+2^{-\frac{1}{2}}+2^{\frac{1}{2}}, giving the answer in the form k\sqrt{2}.

    Factor out 2^{-\frac{3}{2}}\!:\quad 2^{-\frac{3}{2}}\left[1 + 2^3 + 2 + 2^2\right] \;\;=\;\;2^{-\frac{3}{2}}\cdot 15 \;\;=\;\;\frac{15}{2^{\frac{3}{2}}} \;\;=\;\;\frac{15}{2\sqrt{2}}

    Rationalize: . \frac{15}{2\sqrt{2}}\cdot{\color{blue}\frac{\sqrt{  2}}{\sqrt{2}}} \;\;=\;\;\frac{15}{4}\sqrt{2}

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  5. #5
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    Oh, okay!

    I got confused with the rule that (a^2 + b^2) can't be factored, and somehow assumed the expression couldn't be factored.

    Thanks all, for your help!

    P.S. earboth, how did you get \frac{1}{8} \sqrt{2} from \sqrt{\frac{1}{8}}? Shouldn't it be \sqrt{\frac{1}{8}} = \sqrt{2 \times \frac{1}{16}} = \sqrt{\frac{1}{16}}\sqrt{2} = \frac{1}{4}\sqrt{2}?

    Thus \frac{1}{4} \sqrt{2}+2\sqrt{2}+\frac12\sqrt{2}+\sqrt{2} = \sqrt{2}(\frac{1}{4} + 2 + \frac{1}{2} + 1) = \frac{15}{4}\sqrt{2}, which is same as Soroban's answer... yay!
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  6. #6
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    Quote Originally Posted by EvilKitty View Post
    Oh, okay!

    I got confused with the rule that (a^2 + b^2) can't be factored, and somehow assumed the expression couldn't be factored.

    Thanks all, for your help!

    P.S. earboth, how did you get \frac{1}{8} \sqrt{2} from \sqrt{\frac{1}{8}}? Shouldn't it be \sqrt{\frac{1}{8}} = \sqrt{2 \times \frac{1}{16}} = \sqrt{\frac{1}{16}}\sqrt{2} = \frac{1}{4}\sqrt{2}?

    Thus \frac{1}{4} \sqrt{2}+2\sqrt{2}+\frac12\sqrt{2}+\sqrt{2} = \sqrt{2}(\frac{1}{4} + 2 + \frac{1}{2} + 1) = \frac{15}{4}\sqrt{2}, which is same as Soroban's answer... yay!
    Thanks for spotting my mistake. You probably have seen that I forgot to calculate the square-root of \frac14. Sorry for the confusion.
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