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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?

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  • $\begingroup$ Yep, I'm aware of this! I have utilised this before - but doesn't this circumvent the original problem somehow? Suggesting that in some cases it doesn't apply? $\endgroup$ – Charlz97 Jun 8 '18 at 5:38
  • $\begingroup$ It does , that is why i deleted the comment as i has misunderstood the question $\endgroup$ – The Integrator Jun 8 '18 at 5:39
  • $\begingroup$ @Charlz97 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – user Aug 3 '18 at 22:00
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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 $x$.

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  • $\begingroup$ Actually, the actual question I'm currently doing involves integration - the part I surfaced was merely about simplification of the expression that was the integrand. So now..I need to integrate $m sign(x)$ - how do I do it? $\endgroup$ – Charlz97 Jun 8 '18 at 5:41
  • $\begingroup$ @Charlz97 In this case you need to separate the integral in two parts, one for x>0 and ione for x<0. $\endgroup$ – user Jun 8 '18 at 5:43
  • $\begingroup$ And why do my notes ask me to just simplify it as such? Is it because I'm only at high school level? I've no idea what $sign(x)$ is. $\endgroup$ – Charlz97 Jun 8 '18 at 5:43
  • $\begingroup$ @Charlz97 You need to assume m for x>0 and -m for x<0. $\endgroup$ – user Jun 8 '18 at 5:44
  • $\begingroup$ @Charlz97 $sign(x)$ is just a function which gives the sign of $x$, if $x\lt 0$ $sign(x) = -1$ , if $x\gt 0$, $sign(x) = 1$ $\endgroup$ – The Integrator Jun 8 '18 at 5:44
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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 $x>0$.

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  • $\begingroup$ Actually, the actual question I'm currently doing involves integration - the part I surfaced was merely about simplification of the expression that was the integrand. So now..I need to integrate $m sign(x)$ - how do I do it? $\endgroup$ – Charlz97 Jun 8 '18 at 5:41
  • $\begingroup$ And why do my notes ask me to just simplify it as such? Is it because I'm only at high school level? I've no idea what $sign(x)$ is. $\endgroup$ – Charlz97 Jun 8 '18 at 5:43
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    $\begingroup$ @Charlz97 Even at high school level, using the wrong sign in some cases produces wrong results when these cases occur. We might be able to say more with more context $\endgroup$ – Hagen von Eitzen Jun 8 '18 at 5:47
  • $\begingroup$ what is the region you have to integrate over? Because surely, if that area is strictly positive, the sign function will be 1 on the whole area and you can just integrate over $m$ $\endgroup$ – user408856 Jun 8 '18 at 5:50
  • $\begingroup$ $m$ is just a constant, the focus here I think is the sign of $x$. I agree with the rest that more context is needed. It could just be a careless mistake in the notes too. $\endgroup$ – Siong Thye Goh Jun 8 '18 at 5:54

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