What is the square root of x^2?

[tahayassen]

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

 

[arildno]

It is |x|

 

[tahayassen]

It is |x|

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

 

[micromass]

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

No. That is only x if x is positive.

 

[Mark44]

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

 

[haruspex]

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

The square root function (√) is defined (usually) to return the principal square root, i.e. the non-negative one. But the "square root" of x² has two possible values: ±√(x²) = ±|x| = ±x.
As for (x²)^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.

 

[micromass]

As for (x²)^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 x^{1/2} is indeed multivalued. So in that context, you are right.
When working with real number, I do think that the definition of x^{1/2} is the principal square root.
But again, these are just definitions so it's not very interesting.

 

[Mark44]

The square root function (√) is defined (usually) to return the principal square root, i.e. the non-negative one. But the "square root" of x² has two possible values: ±√(x²) = ±|x| = ±x.

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

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 (x²)^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.

 

[haruspex]

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.

 

[Vadar2012]

In high school, I was taught √(x^2) = ±x. It was a long time ago, but will never forget it. Always lost marks for writing anything else.

 

[pwsnafu]

In high school, I was taught √(x^2) = ±x. It was a long time ago, but will never forget it. Always lost marks for writing anything else.

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

 

[arildno]

In high school, I was taught √(x^2) = ±x. It was a long time ago, but will never forget it. Always lost marks for writing anything else.

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...

 

[Mark44]

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 √(x²) has a single value, which depends on whether x is positive or negative.

What you seem to be saying is that √(x²) has two values, ±x. What you said is quoted below.

The square root function (√) is defined (usually) to return the principal square root, i.e. the non-negative one. But the "square root" of x² has two possible values: ±√(x²) = ±|x| = ±x.

An example might help clear this up.

Solve for x: x² = 4

Taking the square root of both sides, we get
√(x²) = √4
|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 exactly the same as the graph of y = √(x).

 

[tahayassen]

Solve for x: x² = 4

Taking the square root of both sides, we get
√(x²) = √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?

 

[arildno]

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?

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

 

[tahayassen]

No, why?

No what?

Edit: -√4 * -√4 = 4

 

[tahayassen]

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!

 

[arildno]

Everything I was taught in high school is wrong!

That is correct.

 

[tahayassen]

{ ({ { x }^{ 2 } }) }^{ \frac { 1 }{ 2 } }=|x|\\ \\ \sqrt { x } =\quad |x|\\ -\sqrt { x } =\quad -|x|

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

edit: Never mind. I'm an idiot.

 

[Mark44]

{ ({ { x }^{ 2 } }) }^{ \frac { 1 }{ 2 } }=|x|\\ \\ \sqrt { x } =\quad |x|\\ -\sqrt { x } =\quad -|x|

You can't go from the first equation to the second. The second line should be
$$ \sqrt { x^2 } = |x| $$

and similarly for the third line.

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

edit: Never mind. I'm an idiot.

 

[HallsofIvy]

In high school, I was taught √(x^2) = ±x. It was a long time ago, but will never forget it. Always lost marks for writing anything else.

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

If the problem were to solve x^2= 5, then the correct answer is \pm\sqrt{5}. Why do you need the "\pm"? Because it is NOT part of the square root. IF \sqrt{5} itself mean both positive and negative values you would NOT need the "\pm".

 

[haruspex]

It seems to me that we are saying different things. What I said (quoted above) is that √(x²) has a single value, which depends on whether x is positive or negative.

What you seem to be saying is that √(x²) has two values, ±x. What you said is quoted below.

Throughout this thread I have written quite consistently that √(x²) = |x|. We are in violent agreement there.
The distinction I'm making is between the square root function (as denoted by the √ symbol), and the generic concept of a square root. The square roots of x² are ±√(x²) = ±|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 values​

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

 

[micromass]

Throughout this thread I have written quite consistently that √(x²) = |x|. We are in violent agreement there.
The distinction I'm making is between the square root function (as denoted by the √ symbol), and the generic concept of a square root. The square roots of x² are ±√(x²) = ±|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 values​

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

 

[tahayassen]

Throughout this thread I have written quite consistently that √(x²) = |x|. We are in violent agreement there.
The distinction I'm making is between the square root function (as denoted by the √ symbol), and the generic concept of a square root. The square roots of x² are ±√(x²) = ±|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 values​

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

 

[arildno]

the disagreement really boils down to if you adhere to the old-fashioned idea of the "square root" as a polyvalued function, or adhere strictly to the idea that "the square root operation" is a single-valued function.
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..

 

[haruspex]

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

One of us does.

 

[Mark44]

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

One of us does.

Where haruspex and I disagreed was in this statement by him in post 6:

But the "square root" of x² has two possible values

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.

Later in that same post, haruspex said this:

As for (x²)^0.5, I would say that does not define a function in the strict sense, so it returns a ± result.

Following this logic, we have
(x²)^1/2 = $\sqrt{x^2}$ = ±x

A reasonable inference from "does not define a function in the strict sense, so it returns a ± result" is that (x²)^0.5 is multi-valued. A graph of the equation y = (x²)^1/2 shows that there is no x value that is mapped to more than one y value.

 

[tahayassen]

Can somebody confirm if I'm doing this right?

{ x }^{ 2 }=4\\ x=|\sqrt { 4 } |\\ x=\pm \sqrt { 4 } \\ x=\pm 2

{ { (x }^{ 2 }) }^{ 0.5 }=4\\ |x|=4\\ x=\pm 4

{ x }^{ 4 }={ 2 }^{ 4 }\\ { x }^{ 2 }=|\sqrt { { 2 }^{ 4 } } |\\ x=|\sqrt { |\sqrt { { 2 }^{ 4 } } | } |

 

[Mark44]

Can somebody confirm if I'm doing this right?

{ x }^{ 2 }=4\\ x=|\sqrt { 4 } |\\ x=\pm \sqrt { 4 } \\ x=\pm 2

Belated edit due to glossing over something you wrote. The second line does not follow from the first.

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]x² = 4
|x| = 2
x = ±2

{ { (x }^{ 2 }) }^{ 0.5 }=4\\ |x|=4\\ x=\pm 4

This is fine.

{ x }^{ 4 }={ 2 }^{ 4 }\\ { x }^{ 2 }=|\sqrt { { 2 }^{ 4 } } |\\ x=|\sqrt { |\sqrt { { 2 }^{ 4 } } | } |

Your answer is needlessly complicated.

x⁴ = 2⁴ = 16
x⁴ - 16 = 0
(x² - 4)(x² + 4) = 0
(x - 2)(x + 2)(x² + 4) = 0
x = 2 or x = -2

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

 

[tahayassen]

How did you go from x⁴ - 16 = 0 to (x² - 4)(x² + 4) = 0? What about the absolute value signs? I know it's a difference of square, but what if was 15 instead of 16?