1. ## Re: Square Root = Two Answers

For all non-negative $\ \ x, \ \ x^{1/2} \ = \ \sqrt{x}.$

$f(x) \ = \ \sqrt{x}$ is a function. As such, it gives one value.

2. ## Re: Square Root = Two Answers

Originally Posted by greg1313
For all non-negative $\ \ x, \ \ x^{1/2} \ = \ \sqrt{x}.$

$f(x) \ = \ \sqrt{x}$ is a function. As such, it gives one value.
that's fine and I agree, but the reason for this is not the reasoning as given on quickmath.com, it's because it is defined this way.

3. ## Re: Square Root = Two Answers

Originally Posted by romsek
that's fine and I agree, but the reason for this is not the reasoning as given on quickmath.com, it's because it is defined this way.
That post didn't state it was from the reasoning as given on quickmath.com. It's a new stand-alone post.

4. ## Re: Square Root = Two Answers

Originally Posted by greg1313
That post didn't state it was from the reasoning as given on quickmath.com. It's a new stand-alone post.
I'm not going to argue with you but that is an interesting interpretation of post #14

5. ## Re: Square Root = Two Answers

Originally Posted by zzephod
It depends on the context. Without qualification I would take it to denote the (two) distinct values of $3e^{n\pi i};\ n\in \mathbb{Z}$, because that is the interpretation in the contexts that I normally work with.
Originally Posted by greg1313
There is no context to consider. You are being asked straight-up for value(s) of $\ \ 9^{1/2}.$
Originally Posted by romsek
I'm not going to argue with you but that is an interesting interpretation of post #14
Given the fact that the original post is is the pre-university elementary algebra forum I find the post #14 spot-on and I find those of zzephod off-putting in that they are totally inappropriate for the level of discussion.
Suppose $x\in\mathbb{R}$ then
$\text{square root(s)}(x)=\begin{cases}\text{does not exist } &: x<0 \\ 0 &: x=0\\ \pm\sqrt x&:x>0\end{cases}$

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