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What is the square root of x^2?

  1. Oct 30, 2012 #1
    It can't be x, because you get a positive number when x is negative.
     
  2. jcsd
  3. Oct 30, 2012 #2

    arildno

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    It is |x|
     
  4. Oct 30, 2012 #3
    What if your have (x^2)^0.5? Doesn't that equal x?
     
  5. Oct 30, 2012 #4
    No. That is only x if x is positive.
     
  6. Oct 30, 2012 #5

    Mark44

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    $$ \sqrt{x^2} = |x|$$
     
  7. Oct 31, 2012 #6

    haruspex

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    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.
     
  8. Oct 31, 2012 #7
    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.
     
  9. Oct 31, 2012 #8

    Mark44

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    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.

     
  10. Nov 1, 2012 #9

    haruspex

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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.
     
  11. Nov 1, 2012 #10
    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.
     
  12. Nov 1, 2012 #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.
     
  13. Nov 1, 2012 #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...
     
  14. Nov 1, 2012 #13

    Mark44

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    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.

    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).
     
  15. Nov 1, 2012 #14
    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?
     
  16. Nov 1, 2012 #15

    arildno

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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
     
  17. Nov 1, 2012 #16
    No what?

    Edit: -√4 * -√4 = 4
     
  18. Nov 1, 2012 #17
    Wow. Everything I was taught in high school is wrong!
     
  19. Nov 1, 2012 #18

    arildno

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    That is correct.
     
  20. Nov 1, 2012 #19
    [tex]{ ({ { x }^{ 2 } }) }^{ \frac { 1 }{ 2 } }=|x|\\ \\ \sqrt { x } =\quad |x|\\ -\sqrt { x } =\quad -|x|[/tex]

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

    edit: Never mind. I'm an idiot.
     
  21. Nov 1, 2012 #20

    Mark44

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    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.
     
  22. Nov 1, 2012 #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]. Why do you need the "[itex]\pm[/itex]"? Because it is NOT part of the square root. IF [itex]\sqrt{5}[/itex] itself mean both positive and negative values you would NOT need the "[itex]\pm[/itex]".
     
    Last edited by a moderator: Nov 2, 2012
  23. Nov 1, 2012 #22

    haruspex

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    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.
     
  24. Nov 1, 2012 #23
    It seems to me that haruspex and Mark44 are saying the exact same thing. But somehow they're not realizing it... :biggrin:
     
  25. Nov 1, 2012 #24
    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.

     
  26. Nov 1, 2012 #25

    arildno

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    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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