Explaining why $\sqrt {x^2+a} = x\sqrt{1+ \frac{a}{x^2}}$ For $x>0$.

Question

I understand the technical operation of extracting $x^2$ out of the root, but is there a way proving it?

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

Answer 1

$$\sqrt{x^2+a}=\sqrt{x^2\left(1+\frac{a}{x^2}\right)}=\underbrace{\sqrt{x^2}}_{=|x|}\cdot \sqrt{1+\frac{a}{x^2}}\underset{(1)}{=}x\sqrt{1+\frac{a}{x^2}}$$

justification:

$(1):$ Because $x>0$.

Answer 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 $|x|=x$.

Answer 3

Note that the right-hand side is positive. Square it and simplify.