# Whats the rule for putting up a plus-minus sign when taking under root?

Given a simple equation....

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

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

$\ x+1 = \pm \surd 21$

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

$a = \pm b$ 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
@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:

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

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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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