Confusion over $\sqrt{ x^2 } = |x|$ for simplifying expression

Question

I know that when solving an equation such as

$\sqrt{ x^2 } = 9$, we must consider $\sqrt{ x^2 } = |x|$

thus $|x|=3$ so $x = 3$ or $-$$x=3$ hence $x = 3$ or $x = -3$

(Of course we usually just go straight to $x = 3$ or $x = -3$ directly.)

What happens, however, when we are simplifying $\sqrt{ x^2 }$ in an expression?

Supposed you are asked to show that an expression $\frac{m}{\sqrt{ x^2 }}$$x$ simplifies to just $m$.

In my notes, it simply goes like this: $\frac{m}{\sqrt{ x^2 }}$$x$ = $\frac{m}{x}$$x$ = m

My question is this, where does $\sqrt{ x^2 } = |x|$ factor in? Conventionally, in simplifying expressions, as compared to solving equations, we seem to just leave out the need for there to be 2 possible solutions (note: no restriction was given at all, no involvement of real-world conditions like given $x > 0$ etc). Why is this so?

Answer 1

Indeed it should be for $x\neq 0$

$$\frac{m}{\sqrt{ x^2 }}x = \frac{m}{|x|}x = sign(x)\cdot m=\begin{cases}m \quad x>0\\\\-m\quad x<0\end{cases}$$

depending upon the sign of $x$.

Answer 2

Assuming $x \ne 0$,$$\frac{m}{\sqrt{x^2}}x=m\frac{x}{|x|}=m\operatorname{sign}(x).$$

The statement that you wrote seems to already imply that $x>0$.