Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
    
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
add comment

4 Answers 4

up vote 1 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 $x$.

share|improve this answer
    
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
add comment

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,

share|improve this answer
    
"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
1  
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
add comment

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).

share|improve this answer
add comment

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:

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.