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Math Help - Simplifying with fractional indeces

  1. #1
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    Simplifying with fractional indeces

    Hello everyone,

    Please could somebody kindly explain what I'm doing wrong here. I have to simplify:

    2^{-{3/2}}+2^{-{1/2}}+2^{{1/2}}+2^{{3/2}}

    My approach so far is:

    =(\frac{1}{\sqrt{2^{3}}}+\frac{1}{\sqrt{2}}+\sqrt{  2}+\sqrt{2^{3}})

    =(\frac{1}{2\sqrt{2}}+\frac{1}{\sqrt{2}}+\sqrt{2}+  2\sqrt{2})

    ... and then I've tried various methods involving simplifying each of the terms, but none of them leads to the correct answer which should be:

    =\bigg(\frac{15}{4}\bigg)\sqrt{2}
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  2. #2
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    Re: Simplifying with fractional indeces

    Quote Originally Posted by mabentley View Post
    Hello everyone,

    Please could somebody kindly explain what I'm doing wrong here. I have to simplify:

    2^{-{3/2}}+2^{-{1/2}}+2^{{1/2}}+2^{{3/2}}

    My approach so far is:

    =(\frac{1}{\sqrt{2^{3}}}+\frac{1}{\sqrt{2}}+\sqrt{  2}+\sqrt{2^{3}})

    =(\frac{1}{2\sqrt{2}}+\frac{1}{\sqrt{2}}+\sqrt{2}+  2\sqrt{2})

    ... and then I've tried various methods involving simplifying each of the terms, but none of them leads to the correct answer which should be:

    =\bigg(\frac{15}{4}\bigg)\sqrt{2}
    I'd suggest multiplying through by the lowest common denominator which is \dfrac{2^{3/2}}{2^{3/2}}

    \dfrac{1 + 2^1 + 2^2 + 2^3}{2^{3/2}} = \dfrac{15}{2^{3/2}} = \dfrac{15}{2\sqrt2}


    Now rationalise the denominator by multiplying top and bottom by \sqrt{2}
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