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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 ?

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Having $\pm$ on the left also doesn't change the outcome. If the left side is "plus", nothing changes. If the left side is "minus", you could just multiply by $-1$ ... – The Chaz 2.0 Jan 3 '12 at 14:15
up vote 2 down vote accepted

$a = \pm b$ 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 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
@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
Thanks again .... perfect ! – explorest Feb 12 '12 at 10:46

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,

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"The $\pm$ is just shorthand." - this, on its own, is the simple answer. :) – J. M. Jan 3 '12 at 14:17
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 3 '12 at 14:21
Oh, the elaboration is dandy. :) I'm merely saying that if one wants to be laconic, that's the short of it. – J. M. Jan 3 '12 at 14:24
Indeed, I agree – Patrick Da Silva Jan 3 '12 at 14:34

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:


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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).

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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:


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


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".

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