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Given a simple equation....

$\ (x+1)^2 =21 $

if we take the under root of both sides , we get

$\ x+1 = \pm \sqrt{21} $

why dont we get a $ \pm $ on the left hand side ?

snoram
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explorest
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8 Answers8

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$a = \pm b$ is shorthand for saying "we don't know what $a$ is, but it has to be either $b$ or $-b$". If you wrote $\pm a = \pm b$ then what you're saying is that

$$ \begin{align} \text{either} & (1) & a&=b \\ \text{or } & (2) & a&=-b \\ \text{or } & (3) & -a&=b \\ \text{or } & (4) & -a&=-b \end{align} $$

But $(1)$ is the same as $(4)$ and $(2)$ is the same as $(3)$, so the first $\pm$ sign is redundant.

You could write $\pm (x+1) = \sqrt{21}$ if you so wished, but it's fairly clear that writing $x+1 = \pm \sqrt{21}$ facilitates solving for $x$.

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    I am grateful for the great explanation....one last thing ... can you please clarify why is square root of 49 = 7 instead of +-7 ?? .... presumably similar reasoning should apply there as well ..... as when we say whats square root of 49 ... the answer could both be +7 and -7 . In other words why don't we entertain that same duality of answer in basic arithmetic ?? – explorest Feb 12 '12 at 10:38
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    @explorest: The confusion here comes from the language rather than the mathematics. When you say "the square root of $49$", you don't really mean what you say: $49$ has two square roots, namely $7$ and $-7$. You could say "the square roots of $49$ are $\pm 7$" and that would be fine; but otherwise saying "the square root of $49$" usually refers to what we write as $\sqrt{49}$. The $\sqrt{\ }$ symbol always refers to the positive root by default, so although $\sqrt{49}=7$ (which is positive) is 'the square root of $49$', $-\sqrt{49}=-7$ is another square root. – Clive Newstead Feb 12 '12 at 10:42
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    Thanks again .... perfect ! – explorest Feb 12 '12 at 10:46
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Really, you should, since $\sqrt{a^2}=|a|$.

You have $$ (x+1)^2=21 $$ which is equivalent to $$ |(x+1)|= \sqrt{21}. $$ Since $|x+1|$ is either $x+1$ or $-(x+1)$ and since $|x+1|=|-(x+1)|$, the above equation is satisfied if and only if either $$\tag{1}x+1=\sqrt{21}\quad\text{or}\quad-(x+1)=\sqrt{21}.$$ This is written in shorthand as:

$$ \pm (x+1)= \sqrt{21}. $$ and read as "$x+1$ is $\sqrt{21}$ or $-(x+1)$ is $\sqrt{21}$".

Now (1) is equivalent to $$\tag{2} x+1=\sqrt{21}\quad\text{or}\quad(x+1)=-\sqrt{21}.$$

And (2) is written in shorthand as $$ (x+1)=\pm\sqrt{21}. $$

This is preferable, since it allows you to solve for $x$ in an expeditious mannar:

$$x=-1\pm\sqrt{21}.$$

David Mitra
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4

You need to understand why we put a $\pm$ sign in the first place. When we say that $$ x^2 = a > 0 \qquad \Longleftrightarrow \qquad x = \pm \sqrt a, $$ It is because we want to say $$ x^2 = a > 0 \qquad \Longleftrightarrow \quad x \in \{ \sqrt a, -\sqrt a \}. $$ The $\pm$ is just a short hand. In other words, when you see a $\pm$ sign, you need to understand that it doesn't mean that "the equation holds whether we put a minus or a plus sign in there", but think of it more as like "the variable on the left-hand side can take on the values of the right-hand side, whether the $\pm$ is actually a $+$ or a $-$.

Hope that helps,

  • "The $\pm$ is just shorthand." - this, on its own, is the simple answer. :) – J. M. ain't a mathematician Jan 03 '12 at 14:17
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    I believe that one who takes the time to ask such a question on MSE deserves more explanation than just "The $\pm$ is just a short hand." because he's probably thought about it longer than $5$ seconds. It's true that it's the key point of the explanation, but I think that detailing is important to understand better. (I am not complaining about your comment, just giving you a feeling of my pedagogical point of view. =D) – Patrick Da Silva Jan 03 '12 at 14:21
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    Oh, the elaboration is dandy. :) I'm merely saying that if one wants to be laconic, that's the short of it. – J. M. ain't a mathematician Jan 03 '12 at 14:24
  • Indeed, I agree – Patrick Da Silva Jan 03 '12 at 14:34
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More interesting is to check out what happens with an inequality:

$$ y^2 \le 4 $$

Since the y-values that satisfy the inequality are between -2 to 2, we write:

$$ |y| \le 2 $$

Without using any step showing plus/minus. In any equation or inequation all we're doing is trying to show which values satisfy it. Thinking of mathematics as a set 'rules' to follow isn't the best way to go about doing it.

Mac
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I'd say some people here are thinking of this backwards. This is how the train of thought should go. Just think about it from the most basic standpoint:

$x^2 =49$, this asks: what value(s) of x equal 49?

We know the answer right away: it could be either 7 or -7 based on normal multiplication with real numbers. Now, this can then be written $x=7,-7$, or $x=\pm7$. Which can then be written $|x|=7$. It can be written as an absolute value because of the definition of absolute value, which is:

$ |x|=\cases{x,\ \ \ \ \text{when} \ x\ge0 \\-x, \ \text{when} \ x<0 }$

Now, let's go back to the beginning, forgetting about the absolute value definition/notation for a minute. Since we know the solutions to our problem have to be either $7$ or $-7$, which can be written as $\pm 7$, then to make the solve and get:

$x=\pm7 \ $from our problem: $x^2=49$

then that means that $\sqrt{x^2}=\pm7= x =\pm \sqrt{49}$

The reason absolute value comes into play is just because of its definition. Because of our example above, combined with the previous given definition of absolute value, mathematicians define:

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

It is a shortcut based on knowledge of multiplication and absolute value definition/notation (that's always true) to write:

$\sqrt{x^2}=|x|=7$

It's also a shortcut to write:

$\sqrt{x^2}=\pm \sqrt{49}$

The reason not to use $\pm$ when no variables are in play is because of two things:

(1): $7^2=49$ or $(-7)^2=49$, aren't a questions at all, they're just true statements.

(2): If you are given a simple expression such as $\sqrt{49}$, standard math notation dictates that it equals $7$. This is just by convention. No $+$ symbol is required and this can be read "positive root of 49" or "principal root of 49".

Zduff
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It's the same whether you take $\pm$ on LHS or RHS. Results obtained using $\pm$ on either side are ultimateley the same. In general, it is customary to write $\pm$ on RHS, since equations are written traditionally in the format LHS=RHS where LHS has argument/variables and RHS has their value.

So, it makes more sense to assign the signs to value rather than variables (in my opinion).

nb1
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I’m surprised not to read any among the answers validating the question, as I would, agreeing that the plus/minus logically belongs on the left. We put it on the right because we can, without changing the evaluation, and because it makes completing the solution simpler from a practical standpoint. But again, logically, the symbol makes the most sense when applied to the left-hand side. Using the x^2=49 example, there is no logical need for the symbol on the right when we take the square root, because the square root of 49 is 7, by definition. But when we take the square root of the left hand side, we must also arrive at a positive number (by definition). In this simple example, we can easily see that a -7 value for x would also satisfies the original equation, so we have to apply a negative to that -7 in order to arrive at the positive (again, by definition) square root. Counter-intuitively, the negative in that “plus or minus” symbol, is to acknowledge that x could represent a negative value, in which case, we’d need the negative of that negative value to arrive at a “square root” that is positive, by definition.

  • Sounds as though you are speaking about the "principal square root". https://mathworld.wolfram.com/PrincipalSquareRoot.html - - - another situation to think about: https://math.stackexchange.com/questions/13801 – 311411 Oct 29 '22 at 14:48
  • Of course (principal square root). – Bernard Oct 29 '22 at 15:10
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Another way to demonstrate the rationale (of it logically belonging on the left-hand side), would be this… Imagine I (on the left) ask you (on the right) to think of a number, an integer, between -10 and 10. Let’s say you think of -7 and I think of x (since I don’t yet know your number). Then I tell you to square it. So you think 49 and I think x^2. I still don’t know your number, but I know we’re still in sync. I tell you to take the square root. You think 7 and I think x. At this point we’re no longer in sync (because x does NOT equal 7, it was -7). But who made the mistake when I said to take the square root, you (on the right) or me (on the left)? It was me; my taking of the square root did not consider that your original number could have been negative. I should have, at that point thought, “if their original number was positive, I could stay in sync with a simple x. But if their original number was negative, in order to stay in sync, I’d have to go with negative x (meaning the opposite of your originally chosen -7). So then I’m thinking -x and you’re thinking 7, which are in sync. However, since I (on the left) don’t KNOW whether you chose a positive or a negative, I (again, on the left) have to think plus-or-minus x in order to cover both potential realities.