# What is the square root of x^2?

by tahayassen
Tags: root, square
 P: 273 $${ ({ { x }^{ 2 } }) }^{ \frac { 1 }{ 2 } }=|x|\\ \\ \sqrt { x } =\quad |x|\\ -\sqrt { x } =\quad -|x|$$ How would I get -|x| from using the exponent method? edit: Never mind. I'm an idiot.
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P: 21,311
 Quote by tahayassen $${ ({ { x }^{ 2 } }) }^{ \frac { 1 }{ 2 } }=|x|\\ \\ \sqrt { x } =\quad |x|\\ -\sqrt { x } =\quad -|x|$$
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.
 Quote by tahayassen How would I get -|x| from using the exponent method? edit: Never mind. I'm an idiot.
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P: 39,565
 Quote by 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.
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 "$\pm$".
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P: 9,873
 Quote by Mark44 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.
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 Quote by haruspex 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 valuesi.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...
P: 273
 Quote by haruspex 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 valuesi.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.

 Quote by arildno 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
 Sci Advisor HW Helper PF Gold P: 12,016 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..
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P: 9,873
 Quote by micromass 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.
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P: 21,311
 Quote by micromass It seems to me that haruspex and Mark44 are saying the exact same thing. But somehow they're not realizing it...
 Quote by haruspex One of us does.
Where haruspex and I disagreed was in this statement by him in post 6:
 Quote by haruspex 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:
 Quote by haruspex 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.
 P: 273 Can somebody confirm if I'm doing this right? $${ x }^{ 2 }=4\\ x=|\sqrt { 4 } |\\ x=\pm \sqrt { 4 } \\ x=\pm 2$$ $${ { (x }^{ 2 }) }^{ 0.5 }=4\\ |x|=4\\ x=\pm 4$$ $${ x }^{ 4 }={ 2 }^{ 4 }\\ { x }^{ 2 }=|\sqrt { { 2 }^{ 4 } } |\\ x=|\sqrt { |\sqrt { { 2 }^{ 4 } } | } |$$
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P: 21,311
 Quote by tahayassen 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.

 Quote by tahayassen
This is fine. Another way that produces the same result is this:x2 = 4
|x| = 2
x = ±2
 Quote by tahayassen $${ { (x }^{ 2 }) }^{ 0.5 }=4\\ |x|=4\\ x=\pm 4$$
This is fine.
 Quote by tahayassen $${ x }^{ 4 }={ 2 }^{ 4 }\\ { x }^{ 2 }=|\sqrt { { 2 }^{ 4 } } |\\ x=|\sqrt { |\sqrt { { 2 }^{ 4 } } | } |$$

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.
 P: 273 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?
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P: 21,311
 Quote by tahayassen How did you go from x4 - 16 = 0 to (x2 - 4)(x2 + 4) = 0?
I factored the expression on the left side.
 Quote by tahayassen What about the absolute value signs? I know it's a difference of square, but what if was 15 instead of 16?
I would do the same thing.

x4 = 15
x4 - 15 = 0
(x2 - √15)(x2 + √15) = 0
(x - ## \sqrt[4]{15}##)(x + ## \sqrt[4]{15}##)(x2 + √15) = 0
x = ## ±\sqrt[4]{15}##

There are also two imaginary roots.
 P: 685 I think the confusion stems from two things: 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: $x = \sqrt{a}$ and $x = -\sqrt{a}$. However, if you are using the square root function, i.e. $\sqrt{x}$, it only returns one value. To make this more clear, define a function $s: X \rightarrow Y$ $s(x) = y$, such that y>0 and y^2=x. This function s returns only one value for an input x. Replace s by the symbol $\sqrt{}$.
 P: 273 Wait a second... Sorry for my repeated lack of understanding, but how did I go from step 2 to step 3 here: $${ x }^{ 2 }=4\\ x=|\sqrt { 4 } |\\ x=\pm \sqrt { 4 } \\ x=\pm 2$$ 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.
 Mentor P: 21,311 The difficulty isn't going from step 2 to step 3 - it's going from step 1 to step 2. x2 = 4 ## \Leftrightarrow## x2 - 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 x2 = 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.
P: 34
 Quote by tahayassen Wait a second... Sorry for my repeated lack of understanding, but how did I go from step 2 to step 3 here: $${ x }^{ 2 }=4\\ x=|\sqrt { 4 } |\\ x=\pm \sqrt { 4 } \\ x=\pm 2$$ 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

$${ x }^{ 2 }=4\\ \sqrt{{ x }^{ 2 }}=\sqrt { 4 } \\ |x| = 2 \\ x=\pm 2$$
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 Quote by Benn I think you might mean $${ x }^{ 2 }=4\\ \sqrt{{ x }^{ 2 }}=\sqrt { 4 } \\ |x| = 2 \\ x=\pm 2$$