Simplifying a radical? How would I simplify the following two radicals.
$$\sqrt{\frac{X^3}{50}}$$ 
For my answer I got $\frac{X^2}{50X}$ but I am not sure if this is correct.
My second question is how would I simplify $\sqrt{\frac{1}{12}}$
 A: No, your simplification is incorrect.
The first radical only makes sense if $x\geq 0$. Assuming this is the case, remember that:


*

*If $a$ and $b$ are both nonnegative, then $\sqrt{ab}=\sqrt{a}\sqrt{b}$, and $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$. 

*$(\sqrt{a})^2 = a$ for any $a\geq 0$.
So:
$$\sqrt{\frac{x^3}{50}} = \frac{\sqrt{x^3}}{\sqrt{50}} = \frac{\sqrt{x^2}\sqrt{x}}{\sqrt{25}\sqrt{2}} = \frac{x\sqrt{x}}{5\sqrt{2}} = \frac{x\sqrt{x}\sqrt{2}}{5\sqrt{2}\sqrt{2}} = \frac{x\sqrt{2x}}{10}.$$
There are other ways of expressing it as well.
For the second, $12 = 4\times 3$, so 
$$\sqrt{12}=\sqrt{4}\sqrt{3} = 2\sqrt{3}.$$
Can you take it from there?
A: For your first one,
$$\sqrt\frac{x^3}{50} = \frac{\sqrt{x^2}\sqrt{x}}{\sqrt{25}\cdot\sqrt{2}} = \frac{x\sqrt{x}}{5\sqrt{2}}$$
At this point, you could multiply top and bottom through by $\sqrt{2}$ (to make the $\sqrt{2}$ part of the bottom "go away"):
$$\frac{x\sqrt{x}\cdot\sqrt{2}}{5\sqrt{2}\cdot\sqrt{2}} = \frac{x\sqrt{2x}}{10}$$
For the second one,
$$\sqrt\frac{1}{12} = \frac{1}{\sqrt{12}} = \frac{1}{\sqrt{4}\cdot\sqrt{3}} = \frac{1}{2\sqrt{3}}$$
From here, you could opt to multiply top and bottom through by $2\sqrt{3}$ as shown below (which is called rationalizing the denominator - to get rid of square root terms in the denominator.)
$$\frac{1\cdot{2\sqrt{3}}}{2\sqrt{3}\cdot{2\sqrt{3}}} = \frac{2\sqrt{3}}{4\cdot\sqrt{9}} = \frac{\sqrt{3}}{2\cdot{3}} = \frac{\sqrt{3}}{6}$$
When dealing with problems like these, it is helpful to split what is under the square root into two separate square root terms (assuming $x$ is nonnegative.) For example,
$$\sqrt\frac{x^3}{50} = \frac{\sqrt{x^3}}{\sqrt{50}}$$
Then, you want to try to rationalize the denominator OR try to manipulate the denominator as I did above to make it more friendly (i.e. $\sqrt{25}\cdot{\sqrt{2}} = \sqrt{50}$, and  since you know that $\sqrt{25} = 5$, $\sqrt{50}$ must equal $5\sqrt{2}$.
