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Say I have $\sqrt{\frac{2}{9}}$. How would I convert this to a mixed radical where the radicand is a whole number?

I can convert the $\sqrt{8}$ to a mixed radical easily, $2\sqrt{2}$, but this fraction is tripping me up.

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Do you have ((the square root of 2) over 8), or (the square root of (2 over 8))? Can you be more specific about the form you want to convert to? –  aschepler Oct 3 '10 at 22:30
I have the square root of (2 over 8), where 2 over 8 is a fraction. I want to convert it into a mixed radical where the radicand is a whole number. –  DMan Oct 3 '10 at 22:31
@number 11: If you accept a rational number factor, you have infinitely many. For instance $\sqrt{\dfrac{2}{8}}=\dfrac{1}{2}=\dfrac{1}{2}\sqrt{1}=\dfrac{1}{4}\sqrt{2}=\ldo‌​ts $ –  Américo Tavares Oct 3 '10 at 22:50
@Américo Tavares- I made a correction (2/9 as opposed to 2/8). My mistake, sorry. –  DMan Oct 3 '10 at 22:52
@number 11: The situation is similar: $\sqrt{\dfrac{2}{9}}=\dfrac{1}{3}\sqrt{2}=\dfrac{1}{6}\sqrt{8}=\ldots $ –  Américo Tavares Oct 3 '10 at 22:56

2 Answers 2

up vote 1 down vote accepted

$\sqrt{\frac{2}{9}} = \frac{\sqrt{2}}{\sqrt{9}} = \frac{\sqrt{2}}{3}$

Is this what you wanted?

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What I wanted is 1/3 sqrt 2, a mixed radical with a whole radicand. –  DMan Oct 4 '10 at 2:11
@DMan: Your respones are quite confusing... You're aware that $(1/3)\sqrt{2}$ is the same thing as $\sqrt{2}/3$, right? –  Hans Lundmark Oct 4 '10 at 6:10
@Hans- Oops, you are right, thanks. –  DMan Oct 4 '10 at 22:08

What do you mean by the radicand is a "whole number"? In your example $\sqrt{8}=2\sqrt{2}$ in both sides of the equation the radicand is a whole number.

In any case, it doesn't really matter because the problem as you stated it doesn't have a radical at all when you simplify: $\sqrt{\frac{2}{8}}=\sqrt{\frac{1}{4}}=\frac{\sqrt{1}}{\sqrt{4}}=\frac{1}{2}$.

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Maybe I'm wrong, but I assumed the first 2 is called the index, and the second was the radicand. Also, sorry for the mistake, I wanted 2/9 as opposed to 2/8. –  DMan Oct 3 '10 at 22:51

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