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I have multiple questions about the $\pm$ sign, since it seems to confuse me in general...

Question 1:

Say I have $15=\pm(a+x)$, Can I use the distributive property so it becomes $15=\pm a \pm x$? Or does that mean I went from 2 to 4 solutions?

I get very confused when I encounter this sign in equations I need to simplify. Whenever I need to deal with these kind of situations I tend to just turn it into two equations, $15= a+x$ and $15=-a-x$. But if I were to do that with bigger equations that still need to be simplified it means I'm wasting alot of time since I'm doing twice the labor...

Question 2:

Does the $\pm$ sign only make sense in equations, or can they be used in normal expressions as well (e.g. $\pm y + 3$). But then I wonder, in what scenario would you do something like this?

Question 3:

$\sqrt {x^2}+37=y+40$

Say I were to simplify $\sqrt{x^2}$ in that equation, I don't know where I should put the $\pm$ sign. Where would I put it? The extra terms 37 and 40 confuse me...

user1534664
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    Your strategy of turning it into two equations is a very sensible one, in my opinion. – TonyK Aug 27 '14 at 13:36

4 Answers4

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If the $\pm$ sign is confusing you, get rid of it. If you have $$15 = \pm(a+x)$$ you can turn that into two equations: $$15 = a+x\\15 = -(a+x)$$ and then deal with the two equations separately, one at a time. That is exactly the meaning of the $\pm$ sign.

The reason you're confused is because the notation is confusing! The expression $\pm a \pm b$ is actually ambiguous: in some contexts it means four values, and in other contexts it means two. Sometimes there is a convention that the two $\pm$ signs must represent the same sign; sometimes there isn't.

In your third example, I'd suggest that you write $z = \sqrt{x^2}$ and turn the equation into $z+37 = y + 40$. Then solve as usual. When you get to the end, you have $z$ and $y$. Then you can conclude that $x$ could be either $z$ or $-z$.

MJD
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  • I like that solution for the third example. Would you think substituting it with z would be better than just changing it to |x|? I'd like to be consistent in which method I use. – user1534664 Aug 27 '14 at 13:44
  • Do whichever you find easiest to think about. – MJD Aug 27 '14 at 13:46
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1: Yes, you can distribute it. Well, it works for all cases. $$\pm(x + y) = \pm x \pm y \quad \forall x,y \in \mathbb R$$

The basic properties are: $$(-)\times (\pm) = (\mp)\\ (\pm)\times(\pm) = (+)\\ (\mp)\times(\mp) = (+)\\ (\pm)\times(\mp) = (-) $$

2: As you said, $\pm$ is used to represent $+$ and $-$ in separate equations. It's a bit more fundamental than that. Even as a statement

$$x \pm y \implies x+ y \text{ and } x - y$$

There isn't a real usefulness for it rather than saves time in writing and speaking. For example,

I ask you, " What is $\sin(A\pm B)$ ?"
You'll tell me, " $\sin A \cos B \pm \cos A \sin B$"

I've recently been using it to help my little brother practice both addition and subtraction simultaneously,

$$3 \pm 2 \to 5, 1\\ 32 \pm 23 \to 55, 9\\ 1729 \pm 999 \to 2728 , 1630\\ 854297992 \pm 299792458 \to 1154090450, 554505534\\ \dots$$ He's been getting really good at it.

3: $\sqrt{x^2} = |x|$

Take the points $\pm x$ on a number line (Hopefully, you understand the usage now) The distance to those points from the origin can be found by the distance formula: $$\sqrt{(\pm x - 0)^2} = \sqrt{x^2}$$

Now, $\sqrt{x^2} = x ,\quad\forall x\in [0, \infty)$
and $\sqrt{x^2} = -x ,\quad\forall x \in (-\infty, 0)$

So, it always outputs the magnitude of $x$

Let's try playing with that equation you've given using our knowledge of this: $$\sqrt {x^2}+37= y + 40 \\ \implies |x| = y + 3 \\ \implies x = y + 3, \space\forall y\in [-3, \infty)\quad\text{ and }\quad x = - (y+3), \space\forall y\in (-\infty, - 3)$$

Nick
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    @TonyK Not necessarily, in some contexts $,\pm a\pm b,$ means $,ua + ub,,$ where $, u = \pm1 .\ $ – Bill Dubuque Aug 27 '14 at 13:38
  • So should I turn it into an absolute value in my equation? e.g. |x| + 37 = y + 40? – user1534664 Aug 27 '14 at 13:39
  • @Nick: the OP might need some explanation. 'Yes' isn't really a hint, I think. – Bart Michels Aug 27 '14 at 13:39
  • Yes. (To convince yourself why this is, plug in a few values. For example: $\displaystyle\sqrt{(3)^2}=\sqrt{9}=3=|3|$, $\displaystyle\sqrt{(0)^2}=\sqrt{0}=0=|0|$, $\displaystyle\sqrt{(-3)^2}=\sqrt{9}=3=|-3|$. (Remember that the $\sqrt{;}$ sign means "the positive root"). – Akiva Weinberger Aug 27 '14 at 13:41
  • @barto: Done! Explanation added. – Nick Aug 27 '14 at 14:25
  • @columbus8myhw: This is something the OP has to convince himself. There are just too many funda-level questions about |x| on this site. But yes, I've added a bitmore explanation! – Nick Aug 27 '14 at 14:29
  • Shouldn't that be $(-x) = y+3$ instead of $x = - (y+3)$? (You've just explained that $|x|$ is either $x$ or $-x$ depending on the sign. So, to make it less confusing, you should have just plugged in $x$ and $-x$ into the equation.) – Akiva Weinberger Aug 27 '14 at 14:32
  • @columbus8myhw: Unless, he's in the 5th grade, I don't think I need to explain why $$-x = a \iff x= -a$$ In due time, I'll keep expanding what I said. – Nick Aug 27 '14 at 17:14
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$\bf{Question\ 1:}$ The plus or minus sign does not distribute like that. To see this, consider the equation $x^{2}-4=0$. It should be obvious that the roots of this equation are $\pm 2$. We can rewrite that as $\pm (10-8)$ since $10-8=2$. However, Distributing the $\pm$ sign through $(10-8)$ would imply that $18$ and $-18$ are also roots of $x^{2}-4=0$.

$\bf{Question\ 2:}$ It can be used as you have specified, although it is not very common to see that.

$\bf{Question\ 3:}$ The quantity $\sqrt{x^{2}}$ is used to denote the magnitude of $x$, which is $|x|$.

graydad
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  • For question 1: Not necessarily. More often than not, if there are two $\pm$ signs in an equation, they have to either be both plus or both minus. (If there is a $\mp$ in the equation, it usually means the opposite sign from whatever the $\pm$'s are.) – Akiva Weinberger Aug 27 '14 at 13:44
  • @columbus8myhw You are right about Question 3, but you are saying exactly what I was trying to say in question 1. ie. There are not four solutions to $\pm(10-8)$, there are only two. It is a good point that more than one set of $pm$ signs can appear in an equation. The signs do need to match when this happens. – graydad Aug 27 '14 at 13:52
  • You're trying to argue that $\pm10\mp8$ implies four solutions, while I'm trying to argue that it can imply two - unless I'm mistaken about what you're trying to say. – Akiva Weinberger Aug 27 '14 at 13:54
  • I was showing why it has to be wrong if you get four solutions, namely because it would mean $18$ and $-18$ are roots to $x^{2}-4=0$ – graydad Aug 27 '14 at 13:56
  • OP was asking if the notation implied four solutions. (The question isn't - is it wrong if I get four solutions? but does writing it this way falsely imply that there are four solutions?) – Akiva Weinberger Aug 27 '14 at 13:59
  • Yes... In my first sentence I say "no, the $\pm$ sign does not distribute like that [in a way that gives four solutions]". I was providing OP an example to show why the $\pm$ sign does not distribute as they initially thought it might. – graydad Aug 27 '14 at 14:02
  • Ah, I see. Sorry, I misinterpreted that first sentence. – Akiva Weinberger Aug 27 '14 at 14:09
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what if variable x = +1 and y = -1, since it is a variable, maybe 4x could be 4+1= 5, not multiplication in this state? y= -1, 2y = 2-1 = 1. 4x -2y=3. 4x-2y=3 _____=3 _

4-2y=4-2y x = 3 ______ 4-2y

3 = +1

4-2y

3 - = +1 3

x=+1 , y=-1

Jet
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  • This could be considered more of a comment . Until you have earned enough reputation to comment please try to leave only clear and concise answers with good explanation and please use MathJax to format answers. welcome to the SE community. – The Integrator Apr 27 '18 at 17:16