# Simplify Fraction with a Square Root?

I'm trying to simplify a problem that I have the answer to but when I check it with Wolfram it seems a bit more simplified and I can't think of what steps are taking place to get that result.

Somehow this: $$\frac{1}{2\sqrt{1-\frac{x^2}{4}}}$$

Becomes this: $$\frac{1}{\sqrt{4-x^2}}$$

And I'm not sure how. I'm guessing its some simple Algebra operation that I'm totally forgetting right now. Any help breaking it down is much appreciated.

We have that $$2=\sqrt{4}$$. Therefore $$2\sqrt{1-\frac{x^2}{4}} = \sqrt{4}\sqrt{1-\frac{x^2}{4}} = \sqrt{4\left(1-\frac{x^2}{4}\right)}=\sqrt{4-x^2}.$$
$$a\sqrt{b} = \sqrt{a^2b}$$.
This moves the 2 inside the $$\sqrt{}$$ where it becomes $$2^2 = 4$$.
We have $$\frac{1}{2\sqrt{1-\frac{x^2}{4}}}=\frac{1}{2\sqrt{\frac{1}{4}(4-x^2)}}$$ $$=\frac{1}{2\sqrt{\frac{1}{4}}\sqrt{4-x^2}}=\frac{1}{\sqrt{4-x^2}}.$$