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Is it mathematically correct to say √16=±4, or would it be correct to say that √16=4, unless specified ±√16? This isn't for any specific problem or anything, but I'm just curious if is correct to have ±, because I know every number has two square roots.

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No it would not be correct to say that the square root of 16 is plus / minus 4.

That is simply because you are using slack wording. "the square root" is singular whereas plus / minus 4 is plural.

Yes the number 16 has two square roots +4 and -4.

What you are thinking about is the square root function y= sqrt(x). Functions are defined to have only one value so we take the positive ones by default and define a second function z=-sqrt(x) to access the engative values.

16, of course is a number, not a function.

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• 2 weeks later...

You're half right studiot. When you are doing basic algebra and solving via "undoing" method, you actually use plus or minus 4. BUT, when you are dealing with coordinates on a function or coefficients of a function you you described, you are correct, because using both roots would not create a function in a Cartesian plane, vertical line test fail.

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In case there's any confusion, the answer to OP's question is that $\sqrt{16}=4$. The notation $\sqrt{x}$, when x is positive, refers to the principal square root, which is also positive. If you want to refer to both square roots of 16, then you should write $\pm \sqrt{16}$. As studiot mentioned, "the square root" usually refers to the principal square root, also.

.

Edited by Dekan
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normally if there is not specification about + or - you should mention it as + or -4.

Or depending on the context, e.g. in basic geometry, one would typically use the principal value, since the negative is intuitively meaningless for describing length, area, etc.

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• 3 weeks later...

Both the forms are correct.

Consider

squareroot(16) = x

16 = x square

so if both - and + comes for x it will be always a positive number

i.e 16 = > (-4) square =16 and 4 (square) = 16

So √16=±4 is correct.

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Both the forms are correct.

Consider

squareroot(16) = x

16 = x square

so if both - and + comes for x it will be always a positive number

i.e 16 = > (-4) square =16 and 4 (square) = 16

So √16=±4 is correct.

Not really.

16 does has two distinct square roots, as does every other real number (besides zero, which has only itself). Nothing will change that. However, notation is not flexible and we denote specific things in specific ways according to convention. The radical $\sqrt{n}$ almost always denotes only the principal value.

The following statements are true according to the typical convention.

$\sqrt{16}=4$

$\sqrt{16}\ne \pm 4$

$\pm\sqrt{16}=\pm 4$

The latter of the three is how one should express both the positive and negative values.

Edited by Amaton
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• 6 months later...

There is no need to put +- in front of root 16 bcoz the no have two roots but roots dosent have to numbers

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Not really.

16 does has two distinct square roots, as does every other real number (besides zero, which has only itself). Nothing will change that. However, notation is not flexible and we denote specific things in specific ways according to convention. The radical $\sqrt{n}$ almost always denotes only the principal value.

The following statements are true according to the typical convention.

$\sqrt{16}=4$

$\sqrt{16}\ne \pm 4$

$\pm\sqrt{16}=\pm 4$

The latter of the three is how one should express both the positive and negative values.

The reason why there are two solutions, the positive and negative, is because if you square a negative number it becomes positive. Therefore, it would be correct to say that $\sqrt{16}= \pm 4$.

Typical convention says that it equals both until specified.

Edited by Unity+
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The notation $\sqrt{x}$, for $x\in\mathbb{R}_{\ge 0}$, refers to the principal square root (which is positive) unless otherwise specified. Amaton's post is correct (except that notation is somewhat flexible--it just depends on how much you want to annoy your readers ).

Edited by John
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The notation $\sqrt{x}$, for $x\in\mathbb{R}_{\ge 0}$, refers to the principal square root (which is positive) unless otherwise specified. Amaton's post is correct (except that notation is somewhat flexible--it just depends on how much you want to annoy your readers ).

True, but I didn't see anything referring to what values were acceptable.

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The notation , for , refers to the principal square root (which is positive) unless otherwise specified. Amaton's post is correct (except that notation is somewhat flexible--it just depends on how much you want to annoy your readers ).

I would dispute that your notation is the same as previously posted.

However the equation (which is not a function) must allow any value that satisfies the equality, by definition.

It is important to distinguish between functions and expressions containing an equals sign.

When we have an expression containing an equals sign that may be satisfied by more than one value we would normally have additional conditions to satisfy which will determine which one we want.

There were several mistakes mades in a recent thread on quadratics by posters failing to observe this simple requirement.

So John's statement is a refers to a set of values for x (the domain) and another set of values that picks out a unique member for each value of x.

That properly makes it a function.

Amaton referred to a specific value of x (16) and presented it via an equals sign.

That does not qualify as a function.

Edited by studiot
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There are two square roots of 16, but $\sqrt{16} = 4$ unless otherwise specified. In general, $\forall x \in \mathbb{R}, \sqrt{x^{2}} = |x|$. If we have the equation $x^{2} - 16 = 0$, then $x = \pm \sqrt{16} = \pm 4$. It's not a huge deal, regardless, but Amaton's post is correct.

Edited by John
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There are two square roots of 16, but $\sqrt{16} = 4$ unless otherwise specified. In general, $\forall x \in \mathbb{R}, \sqrt{x^{2}} = |x|$. If we have the equation $x^{2} - 16 = 0$, then $x = \pm \sqrt{16} = \pm 4$. It's not a huge deal, regardless, but Amaton's post is correct.

I didn't agree with you before and I don't agree with you now.

Take for instance the equation x2+3x = -2

Would you bar the negative roots?

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Er, no, why? Finding the roots of a general polynomial equation isn't the same as taking the square root of a real number. And as stated before, if you look at $x^{2} - 16 = 0$, then there are two solutions, $\sqrt{16} = 4$ and $-\sqrt{16} = -4$. But the notation, $\sqrt{16}$, refers to the principal square root of 16, i.e. 4. Disagree if you like, but that is the convention.

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I don't agree with your convention and you still haven't addressed my point that this is because you are talking about a function not an equation.

What would you make of this square root ?

$\sqrt {\left( {5 + \sqrt {21} } \right)}$

Which again is not a function

or perhaps this one, which is

In particular can you guarantee that it is always positive ?

$\sqrt {\left( {1 - 2x + 5{x^2} - 4{x^3} + 4{x^4}} \right)}$

Here are some more thoughts.

Do you consider

$\sqrt {{{\left( x \right)}^2}}$$= x$

To be an identity?

If so do you agree that

$\sqrt {{{\left( { - 4} \right)}^2}}$$=$$- 4$

Edited by studiot
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It's not my convention. It's the convention used by mathematicians in general. While the fact that it's a convention means there's probably no absolute authority to make the definition, a Google search for definitions of the square root or the radical sign should convince you.

And I'm honestly not sure what your point is. All I've said is that $\forall x \in \mathbb{R}, \sqrt{x^{2}} = |x|$. It's simply a matter of notation. Questions about the roots of various polynomial equations are somewhat unrelated.

But for the particular example of $\sqrt {\left( {1 - 2x + 5{x^2} - 4{x^3} + 4{x^4}} \right)}$, so long as $1 - 2x + 5{x^2} - 4{x^3} + 4{x^4} > 0$, then yes, $\sqrt {\left( {1 - 2x + 5{x^2} - 4{x^3} + 4{x^4}} \right)} > 0$.

Using Wolfram Alpha, it seems the two square roots of $1 - 2x + 5{x^2} - 4{x^3} + 4{x^4}$ (which, as it turns out, is positive for all real x) are $\pm(2x^2 - x + 1)$, so $\sqrt {\left( {1 - 2x + 5{x^2} - 4{x^3} + 4{x^4}} \right)} = |2x^2 - x + 1|$, which is positive.

Edit: Just saw your edit. The notation indicates the principal square root, so $\sqrt{(-4)^{2}} = |{-4}| = 4$.

Edited by John
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John post#18

And I'm honestly not sure what your point is.

Perhaps that's because you have still not acknowledged what I said several times about the difference between functions and equations (or expressions containg equations).

Studiot post#13

It is important to distinguish between functions and expressions containing an equals sign.

Incidentally I'm sorry the material I added whilst you were thinking was tagged onto my previous post by the system. I genuinely tried to make a new post.

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The difference between an equation and a function is irrelevant. The notation $\sqrt{x}$ refers to the principal square root of x. It's not strange. It's not controversial. I don't know why this is such a sticking point.

Edited by John
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The difference between an equation and a function is irrelevant.

Why do you say this?

IMHO it is of the essence of the issue.

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We're just talking about the definition of the radical sign. By default, it indicates the principal (i.e. positive) square root of the radicand. In situations where both square roots of the radicand are desired, we prepend $\pm$ to specify this. So if someone asks for $\sqrt{16}$, i.e. "the square root of 16," the answer is 4. If someone asks for solutions to the equation $x^2 = 16$, i.e. "what numbers are square roots of 16," the answers are 4 and -4, and we can express these as $\pm\sqrt{16}$. If you want to use $\sqrt{16}$ to mean both 4 and -4, then that's fine, but it may cause some confusion unless this usage is specified (though often the intended meaning will probably be clear from the context).

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You keep making statements as absolute, without any backing or logical development.

For instance you keep appealing to mathematical convention.

Here is an excerpt from the most successful and long lived textbook on the subject ever written in English (and translated into umpteen languages) which has survived for more than a century as the gold standard

116 Definition The root of any proposed expression is that quantity which being multiplied by itself the requisite number of times produces the given expression.

The operation of finding the root is called Evolution. It is the reverse of Involution.

117 By the Rule of signs we see that

(1) any even root of a positive quantity may be either positive or negative;

!2) no negative quantity can have an even root

(3)every odd root of a quantity has the same sign as the quantity itself

Has someone recently proved messers Hall & Knight wrong in that last century?

I have not heard of it.

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I'm stating the (usual) definition of notation. You might as well ask me to logically develop why we normally use "+" to denote addition. If you did, then I'd say the same thing: It's a matter of convention, and a Google search should provide sufficient evidence to show this.

Notation isn't sacred. If I like, I can define $\sqrt{x}$ to refer to $e^{x}$ or $x \choose {x-6}$ or anything else. It'd be odd, and there's no good reason--or at least, no reason that outweighs the confusion it'd likely cause--to do so, but I could. However, the usual definition of $\sqrt{x}$ is the principal square root of x. This doesn't require rigorous development. It's simply the way notation has developed over the years.

This definition also doesn't conflict with the quoted passage, unless you're referring to the fact that I translated $\sqrt{x}$ as "the square root of x." But if that bothers you, then ignore it. The rest of my post remains true.

Edited by John
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Well actually it said any even root may be positive or negative. Does that not include the square root?

Edited by studiot

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