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?
$$\sqrt {x^2+a} = x\sqrt{1+ \frac{a}{x^2}}$$

Assuming that $x>0$ we have that
\begin{align}
\sqrt{x^2+a} &=\\
\sqrt{x^2\left(1+\frac{a}{x^2}\right)} &=\\
\sqrt{x^2}\sqrt{1+\frac{a}{x^2}} &=\\
|x| \sqrt{1+\frac{a}{x^2}}&=\\
x\sqrt{1+\frac{a}{x^2}}.
\end{align}
In the first equality we must require that $x\neq 0$, since otherwise the division by $x^2$ is not defined. For the next equality we have used that $\sqrt{m\cdot n} = \sqrt{m}\sqrt{n}$, which holds whenever $m,n\geq 0$. For the last equality we have, in the general case, that $\sqrt{x^2} = |x|$. But since we assumed that $x>0$ we have that $|x|=x$.
