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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

    micromass

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    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

    micromass

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    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.
     
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