Explaining why $\sqrt {x^2+a} = x\sqrt{1+ \frac{a}{x^2}}$ For $x>0$. I understand the technical operation of extracting $x^2$ out of the root, but is there a way proving it?
$$\sqrt {x^2+a} = x\sqrt{1+ \frac{a}{x^2}}$$

$$\sqrt{x^2+a}=\sqrt{x^2\left(1+\frac{a}{x^2}\right)}=\underbrace{\sqrt{x^2}}_{=|x|}\cdot \sqrt{1+\frac{a}{x^2}}\underset{(1)}{=}x\sqrt{1+\frac{a}{x^2}}$$
justification:
$(1):$ Because $x>0$.
