Radicals Do they always have a +/- in front of them?

[lLovePhysics]

Do radicals always have a +/- sign in front of them?

For example the equation: $$y=\sqrt{x}$$

Is that only a half of a sleeping parabola or is it a full one?

Does a radical only have a +/- sign in front of it when it takes this form?:

$$y^{2}=x \rightarrow y=\pm \sqrt{x}$$


So, is there a rule that when you square root something the +/- sign always goes in front of it? If you do not square root something, the square rooted term is only positive and does not contain the +/- sign?

Thanks in advance.

 

[quasar987]

The ± pops up when you have an equation with an unknown squared on one side and you want to solve for that unknown. Then you take the square root of both sides and the ± pops up because if we have for instance y² = 9, then both y=3 and y=-3 are solutions. So we write $y^2=x \ \Leftrightarrow \ y=\pm \sqrt{x}$ to illustrate that fact.

If you have the function $y=\sqrt{x}$, then this is quite different from what I talk about above. It is an equation that simply tells you "given a number x in the domain of definition of the function y, associate to it the number $\sqrt{x}$".

 

[lLovePhysics]

The ± pops up when you have an equation with an unknown squared on one side and you want to solve for that unknown. Then you take the square root of both sides and the ± pops up because if we have for instance y² = 9, then both y=3 and y=-3 are solutions. So we write $y^2=x \ \Leftrightarrow \ y=\pm \sqrt{x}$ to illustrate that fact.

If you have the function $y=\sqrt{x}$, then this is quite different from what I talk about above. It is an equation that simply tells you "given a number x in the domain of definition of the function y, associate to it the number $\sqrt{x}$".

So if you are given the equation $$y=\sqrt{x}$$, but it doesn't tell you that it is a function should you always assume that it is a function and that the sign in front of the radical is positive?

What if it tells you to test whehter $$y=\sqrt{x}$$ is a function? Should you still assume that it is always a function? and always positive?

 

[quasar987]

there is nothing to assume regarding the sign in front of the radical in $$y=\sqrt{x}$$: $$\sqrt{x}=+\sqrt{x}$$.

 

[HallsofIvy]

$y= \sqrt{x}$ is, by definition, the positive root. That's why you need the $\pm$ in front of a radical in the solution of an equation. The complete solution to x²= 3 is $x=\pm \sqrt{3}$. If you only wrote $x= \sqrt{3}$ you would be missing the negative root. Yes, the graph of $y= \sqrt{x}$ is the upper half of a parabola with horizontal axis. Notice that x= y² would not define y as a function of x.

 

[Gib Z]

Is that only a half of a sleeping parabola or is it a full one?

That is so Cute! (Stop looking at me like that).

You had the right answer in your post :) (Although your arrow could point both ways, it still works in reverse)

$$y^{2}=x \rightarrow y=\pm \sqrt{x}$$

But when $$y=\sqrt{x}$$ we always take the positive root, its convention.

 

[symbolipoint]

$$y=\sqrt{x}$$

is a different function than

$$y=\pm\sqrt{x}$$
or do I mean $$y=\-sqrt{x}$$. having trouble using LaTex here.
I'm trying to say y=-(x)^(0.5)

 

[HallsofIvy]

$y= \sqrt{x}$ is a "function" while "$y= \pm\sqrt{x}$ is not a function. If you meant $y= -\sqrt{x}$, yes, that is a different function from $y= \sqrt{x}$, just as y= -x is different from y= x.

 

[Feldoh]

You add +/- if you yourself added the radical to the problem. Take your example:

$$y^2 = x$$

"y" could be both positive or negative since it is being squared right? We account for this by adding +/-.

Don't confuse this with a function that is defined with a radical such as:

$$y = \sqrt{x}$$

Where x can only be positive.