Simplifying with fractional indeces

• Sep 15th 2011, 10:52 AM
mabentley
Simplifying with fractional indeces
Hello everyone,

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

$\displaystyle 2^{-{3/2}}+2^{-{1/2}}+2^{{1/2}}+2^{{3/2}}$

My approach so far is:

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

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

... 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:

$\displaystyle =\bigg(\frac{15}{4}\bigg)\sqrt{2}$
• Sep 15th 2011, 11:12 AM
e^(i*pi)
Re: Simplifying with fractional indeces
Quote:

Originally Posted by mabentley
Hello everyone,

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

$\displaystyle 2^{-{3/2}}+2^{-{1/2}}+2^{{1/2}}+2^{{3/2}}$

My approach so far is:

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

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

... 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:

$\displaystyle =\bigg(\frac{15}{4}\bigg)\sqrt{2}$

I'd suggest multiplying through by the lowest common denominator which is $\displaystyle \dfrac{2^{3/2}}{2^{3/2}}$

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

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