- #1

tahayassen

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It can't be x, because you get a positive number when x is negative.

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- Thread starter tahayassen
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- #1

tahayassen

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It can't be x, because you get a positive number when x is negative.

- #2

arildno

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It is |x|

- #3

tahayassen

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It is |x|

What if your have (x^2)^0.5? Doesn't that equal x?

- #4

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What if your have (x^2)^0.5? Doesn't that equal x?

No. That is only x if x is positive.

- #5

Mark44

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$$ \sqrt{x^2} = |x|$$

- #6

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The square rootIt can't be x, because you get a positive number when x is negative.

As for (x

- #7

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As for (x^{2})^{0.5}, I would say that does not define a function in the strict sense, so it returns a ± result. Prepared to be shouted down on that one, though.

I guess it depends on the definition of "exponentiation". In complex analysis, the operation [itex]x^{1/2}[/itex] is indeed multivalued. So in that context, you are right.

When working with real number, I do think that the definition of [itex]x^{1/2}[/itex] is the principal square root.

But again, these are just definitions so it's not very interesting.

- #8

Mark44

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This is not true. ##\sqrt{x^2} = |x|##.The square rootfunction(√) is defined (usually) to return the principal square root, i.e. the non-negative one. But the "square root" of x^{2}has two possible values: ±√(x^{2}) = ±|x| = ±x.

Can you cite a single resource that makes the claim that, for example √4 = -2?

Although we say that every positive number has two square roots, a positive one and a negative, when we refer to "the square root" of something, we're talking about the principal (or positive) square root.

As for (x^{2})^{0.5}, I would say that does not define a function in the strict sense, so it returns a ± result. Prepared to be shouted down on that one, though.

- #9

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That's what I said. The distinction I'm drawing is between taking a square root in a calculation, in which you have to allow for both signs, and the definition of the square root function (as indicated by √), which must be single valued by definition.This is not true. ##\sqrt{x^2} = |x|##.

- #10

Vadar2012

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- #11

pwsnafu

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Ouch that sucks. Vibes.

Your teacher should have known better: if it was equal to ±x, the quadratic formula wouldn't need it.

- #12

arildno

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High schools all over the world are, for some reason, terrified of the absolute value sign, and thus teaches wrongly.

I do not know why they are so scared...

- #13

Mark44

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This is not true. ##\sqrt{x^2} = |x|##.

That's what I said. The distinction I'm drawing is between taking a square root in a calculation, in which you have to allow for both signs, and the definition of the square root function (as indicated by √), which must be single valued by definition.

It seems to me that we are saying different things. What I said (quoted above) is that √(x

What you seem to be saying is that √(x

The square rootfunction(√) is defined (usually) to return the principal square root, i.e. the non-negative one. But the "square root" of x^{2}has two possible values: ±√(x^{2}) = ±|x| = ±x.

An example might help clear this up.

Solve for x: x

Taking the square root of both sides, we get

√(x

|x| = 2

If x < 0, then x = -2

If x > 0, then x = 2

A fine point here is that |x| ≠ ±x. If that were the case, then the graph of y = |x| would not represent a function, since each nonzero x value would have two y values. The resulting graph would be the combined graphs of y = x and y = -x.

Instead, the graph of y = |x| has a V shape, and is

- #14

tahayassen

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Solve for x: x^{2}= 4

Taking the square root of both sides, we get

√(x^{2}) = √4

|x| = 2

Why did you only take the positive square-root of both sides? Shouldn't you take both the positive and negative square-roots of both sides since both are valid solutions?

- #15

arildno

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No, why?Why did you only take the positive square-root of both sides? Shouldn't you take both the positive and negative square-roots of both sides since both are valid solutions?

Taking the NEGATIVE square root operation is another, EQUALLY valid operation, but with absolutely no new information gained.

Your answer would then be:

-|x|=-2

- #16

tahayassen

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- #17

tahayassen

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No, why?

Taking the NEGATIVE square root operation is another, EQUALLY valid operation, but with absolutely no new information gained.

Your answer would then be:

-|x|=-2

Wow. Everything I was taught in high school is wrong!

- #18

arildno

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Everything I was taught in high school is wrong!

That is correct.

- #19

tahayassen

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How would I get -|x| from using the exponent method?

edit: Never mind. I'm an idiot.

- #20

Mark44

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You can't go from the first equation to the second. The second line should be[tex]{ ({ { x }^{ 2 } }) }^{ \frac { 1 }{ 2 } }=|x|\\ \\ \sqrt { x } =\quad |x|\\ -\sqrt { x } =\quad -|x|[/tex]

$$ \sqrt { x^2 } = |x| $$

and similarly for the third line.

How would I get -|x| from using the exponent method?

edit: Never mind. I'm an idiot.

- #21

HallsofIvy

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With your teacher not here to defend him or herself, I'm going to take that with a grain of salt. I strongly suspect you misunderstood your teacher.

If the problem were to solve the equation, [itex]x^2= 4[/itex], then the correct answer would be [itex]\pm 2[/itex]. If the problem were to find [itex]\sqrt{4}[/itex] then the correct answer is [itex]2[/itex] NOT "[itex]\pm 2[/itex]".

If the problem were to solve [itex]x^2= 5[/itex], then the correct answer is [itex]\pm\sqrt{5}[/itex].

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- #22

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Throughout this thread I have written quite consistently that √(xIt seems to me that we are saying different things. What I said (quoted above) is that √(x^{2}) has a single value, which depends on whether x is positive or negative.

What you seem to be saying is that √(x^{2}) has two values, ±x. What you said is quoted below.

The distinction I'm making is between the square root

The point of disagreement is extremely subtle: the use of the definite article. I wrote

the "square root" has two possible values

i.e. in the generic sense of square root; you prefer to reserve "- #23

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Throughout this thread I have written quite consistently that √(x^{2}) = |x|. We are in violent agreement there.

The distinction I'm making is between the square rootfunction(as denoted by the √ symbol), and the generic concept of a square root. The square roots of x^{2}are ±√(x^{2}) = ±|x|, which is the same as ±x.

The point of disagreement is extremely subtle: the use of the definite article. I wrote

the "square root" has two possible valuesi.e. in the generic sense of square root; you prefer to reserve "thesquare root" to mean the principal square root. Fair enough, but I think that's a matter of taste, and I'm not sure how else I could have worded it. "A square root has two possible vales"? No. "Taking the square root produces two possible values"? Still that definite article.

It seems to me that haruspex and Mark44 are saying the exact same thing. But somehow they're not realizing it...

- #24

tahayassen

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Throughout this thread I have written quite consistently that √(x^{2}) = |x|. We are in violent agreement there.

The distinction I'm making is between the square rootfunction(as denoted by the √ symbol), and the generic concept of a square root. The square roots of x^{2}are ±√(x^{2}) = ±|x|, which is the same as ±x.

The point of disagreement is extremely subtle: the use of the definite article. I wrote

the "square root" has two possible valuesi.e. in the generic sense of square root; you prefer to reserve "thesquare root" to mean the principal square root. Fair enough, but I think that's a matter of taste, and I'm not sure how else I could have worded it. "A square root has two possible vales"? No. "Taking the square root produces two possible values"? Still that definite article.

I learned in this thread from arildno that taking both the positive and negative square roots of x is actually pointless because it's doesn't give you any additional information. Not disagreeing with you or anything. Just saying that the argument between you two is a moot point.

No, why?

Taking the NEGATIVE square root operation is another, EQUALLY valid operation, but with absolutely no new information gained.

Your answer would then be:

-|x|=-2

- #25

arildno

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The idea of polyvalued functions is really redundant, since you can define one-valued function to the power set of R, or any other set.

but, I'm digressing here..

- #26

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One of us does.It seems to me that haruspex and Mark44 are saying the exact same thing. But somehow they're not realizing it...

- #27

Mark44

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micromass said:It seems to me that haruspex and Mark44 are saying the exact same thing. But somehow they're not realizing it...

Where haruspex and I disagreed was in this statement by him in post 6:haruspex said:One of us does.

IMO, this statement is not as clear as it could be, as it does not seem to exclude the possibility of both values occurring simultaneously. That was the heart of our disagreement.haruspex said:But the "square root" of x^{2}has two possible values

Later in that same post, haruspex said this:

Following this logic, we haveharuspex said:As for (x^{2})^{0.5}, I would say that does not define a function in the strict sense, so it returns a ± result.

(x

A reasonable inference from "does not define a function in the strict sense, so it returns a ± result" is that (x

- #28

tahayassen

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[tex]{ x }^{ 2 }=4\\ x=|\sqrt { 4 } |\\ x=\pm \sqrt { 4 } \\ x=\pm 2[/tex]

[tex]{ { (x }^{ 2 }) }^{ 0.5 }=4\\ |x|=4\\ x=\pm 4[/tex]

[tex]{ x }^{ 4 }={ 2 }^{ 4 }\\ { x }^{ 2 }=|\sqrt { { 2 }^{ 4 } } |\\ x=|\sqrt { |\sqrt { { 2 }^{ 4 } } | } |[/tex]

- #29

Mark44

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Belated edit due to glossing over something you wrote. The second line does not follow from the first.Can somebody confirm if I'm doing this right?

[tex]{ x }^{ 2 }=4\\ x=|\sqrt { 4 } |\\ x=\pm \sqrt { 4 } \\ x=\pm 2[/tex]

In the first equation, there are two solutions. In the second equation, there is only one solution, since |√4| is the same as √4, which is 2.

[STRIKE]This is fine. Another way that produces the same result is this:[/STRIKE]x

|x| = 2

x = ±2

This is fine.[tex]{ { (x }^{ 2 }) }^{ 0.5 }=4\\ |x|=4\\ x=\pm 4[/tex]

Your answer is needlessly complicated.[tex]{ x }^{ 4 }={ 2 }^{ 4 }\\ { x }^{ 2 }=|\sqrt { { 2 }^{ 4 } } |\\ x=|\sqrt { |\sqrt { { 2 }^{ 4 } } | } |[/tex]

x

x

(x

(x - 2)(x + 2)(x

x = 2 or x = -2

There are also two imaginary solutions - x = 2i and x = -2i.

Last edited:

- #30

tahayassen

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- #31

Mark44

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I factored the expression on the left side.How did you go from x^{4}- 16 = 0 to (x^{2}- 4)(x^{2}+ 4) = 0?

I would do the same thing.What about the absolute value signs? I know it's a difference of square, but what if was 15 instead of 16?

x

x

(x

(x - ## \sqrt[4]{15}##)(x + ## \sqrt[4]{15}##)(x

x = ## ±\sqrt[4]{15}##

There are also two imaginary roots.

- #32

Edgardo

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1. Solving for x in the equation x^2 = a.

2. Using the square root function.

If you are given the equation x^2 = a, then there are two solutions, namely a positive and a negative: [itex]x = \sqrt{a}[/itex] and [itex]x = -\sqrt{a}[/itex].

However, if you are using the square root function, i.e. [itex]\sqrt{x}[/itex], it only returns one value. To make this more clear, define a function

[itex]s: X \rightarrow Y[/itex]

[itex]s(x) = y[/itex], such that y>0 and y^2=x.

This function s returns only one value for an input x. Replace s by the symbol [itex]\sqrt{}[/itex].

- #33

tahayassen

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[tex]{ x }^{ 2 }=4\\ x=|\sqrt { 4 } |\\ x=\pm \sqrt { 4 } \\ x=\pm 2[/tex]

If I remember correctly, to take the absolute value of anything, you just ensure that the sign is positive. For example, |-2|=2, |4|=4, |-sqrt(4)|=sqrt(4), and |sqrt(4)|=4.

- #34

Mark44

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The difficulty isn't going from step 2 to step 3 - it's going from step 1 to step 2.

x^{2} = 4

## \Leftrightarrow## x^{2} - 4 = 0

## \Leftrightarrow## (x - 2)(x + 2) = 0

## \Leftrightarrow## x = ± 2

|√4| = √4 = 2, not ±2.

BTW, tahayassen, you had essentially the same set of equations in post #28, a week ago. I missed that you had gone from

x^{2} = 4 to

x = |√4|.

The problem here is that the first equation has two solutions, while the second equation has only one.

I have gone back and edited my post.

x

## \Leftrightarrow## x

## \Leftrightarrow## (x - 2)(x + 2) = 0

## \Leftrightarrow## x = ± 2

|√4| = √4 = 2, not ±2.

BTW, tahayassen, you had essentially the same set of equations in post #28, a week ago. I missed that you had gone from

x

x = |√4|.

The problem here is that the first equation has two solutions, while the second equation has only one.

I have gone back and edited my post.

Last edited:

- #35

Benn

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[tex]{ x }^{ 2 }=4\\ x=|\sqrt { 4 } |\\ x=\pm \sqrt { 4 } \\ x=\pm 2[/tex]

If I remember correctly, to take the absolute value of anything, you just ensure that the sign is positive. For example, |-2|=2, |4|=4, |-sqrt(4)|=sqrt(4), and |sqrt(4)|=4.

I think you might mean

[tex]{ x }^{ 2 }=4\\ \sqrt{{ x }^{ 2 }}=\sqrt { 4 } \\ |x| = 2 \\ x=\pm 2[/tex]

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