Combining a radical and simplifying? How would I combine and simplify the following radical:
$$\sqrt {\frac{A^2}{2}} - \sqrt \frac{A^2}{8}$$
 A: $$\sqrt {\frac{A^2}{2}} - \sqrt \frac{A^2}{8}\\=\frac{|A|}{\sqrt 2}-\frac{|A|}{2\sqrt 2}\\=\frac{2|A|-|A|}{2\sqrt 2}\\=\frac{|A|}{2\sqrt 2}\frac{\sqrt 2}{\sqrt 2}\\=\frac{\sqrt 2|A|}{4}$$
A: $\sqrt{\frac{A^2}{2}} - \sqrt{\frac{A^2}{8}} = \frac{A}{\sqrt{2}} - \frac{A}{2\sqrt{2}} = \frac{2A}{2\sqrt{2}} - \frac{A}{2\sqrt{2}} = \frac{A}{2\sqrt{2}}$
Assuming $A \geq 0$. If $A < 0$, you can replace with $|A|$. 
A: Recall the following facts.
$$\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}} \text{ whenever }a,b>0$$
$$\sqrt{x^2} = \lvert x \rvert \text{, where $x \in \mathbb{R}$}$$
$$\sqrt{ab} = \sqrt{a} \sqrt{b} \text{ whenever }a,b>0$$
Using the above, we then get that
\begin{align}
\sqrt{\dfrac{A^2}{2}} - \sqrt{\dfrac{A^2}{8}} & = \sqrt{\dfrac{A^2}{2}} - \sqrt{\dfrac{A^2}{2^2 \times 2}} = \dfrac{\sqrt{A^2}}{\sqrt{2}} - \dfrac{\sqrt{A^2}}{\sqrt{2^2 \times2}}\\
& = \dfrac{\lvert A\rvert}{\sqrt{2}} - \dfrac{\lvert A \rvert}{\sqrt{2^2} \times \sqrt{2}} = \dfrac{\lvert A\rvert}{\sqrt{2}} - \dfrac{\lvert A \rvert}{2\sqrt{2}}\\
& =\dfrac{\lvert A \rvert}{2\sqrt{2}}
\end{align}
