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

1. ## 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?

2. ## 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.

3. Originally Posted by EvilKitty
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?

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

5. 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!

6. Originally Posted by EvilKitty
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.