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Why is $4^{1/2}=+2$? It should also be $-2$ since both squared just give two only. Also why do we always represent root of $x$ on the right side of the number line?

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marked as duplicate by tomasz, Pedro Tamaroff, Ross Millikan, Brian M. Scott, Zev Chonoles Jun 1 '13 at 0:09

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It is an Agreement: in the real case, we always take the positive root. Among other reasons, to avoid confusion and to make $\,f(x)=\sqrt x\,\,,\,\,x\ge 0\,$ an actual function. – DonAntonio May 31 '13 at 23:40
The square root function has two branches. You can consider the positive one or the negative one. So if you want negative values you should write $f(x)=-\sqrt{x}$. – Sigur May 31 '13 at 23:41
You are confused with the idea of solving the equation x²=4 which is indeed 2,-2 Think about taking a squareroot as performing an operation on a number that gives uniquely another number. – imranfat May 31 '13 at 23:42
@ex0du5, I didn't say such a thing. I implied that if we do not agree on what real value to take in the real square root, two people working with the function "square root", assuming it is already a function, can get pretty different values, even if restricted to real numbers. Now, somebody already mentioned branches and stuff, which seems to be way over the level of somebody asking such a basic question. – DonAntonio Jun 1 '13 at 0:08
Many high school teachers would write $\sqrt{4}=\pm 2$. They are not really wrong. Just speaking a different language. – André Nicolas Jun 1 '13 at 1:07
up vote 5 down vote accepted

It is by convention: with real numbers, we agree to take the positive square root. This allows us to define $$f(x) = \sqrt x, \;\;x \in \mathbb R, \;\;x\geq 0, $$ so it is a true real-valued function: taking a square-root of a number greater than or equal to zero "returns" a unique real number (is hence a function). Without that convention $\sqrt x$ would fail to be be a function.

(Note: as imranfat suggests: I think you might be confusing the square root function with what we know about solving an equation $x^2 = 4$, which has two solutions, $x = \pm 2$.

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