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Is the square root of 4 a constant?

  1. Apr 9, 2015 #1
    I don't know what to say. sqrt(4) = +/- 2, but constants are fixed. When written sqrt(4), it seems sqrt(4) is a constant, but not in the latter case. Help!
     
  2. jcsd
  3. Apr 9, 2015 #2

    micromass

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    The square root is by definition positive. So ##\sqrt{4} = 2##.
     
  4. Apr 9, 2015 #3

    HallsofIvy

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    You are misunderstanding the notation. There are two numbers, -2 and 2, whose square is 4. But only one of those, 2, is "the square root of four".
    And, yes, both "2" and "-2" are "constant".
     
  5. Apr 9, 2015 #4
    But (-2)^2 = 4?
     
  6. Apr 9, 2015 #5

    russ_watters

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    Yes. Those are different statements though.
     
  7. Apr 9, 2015 #6
    Maybe one way of looking at it is to ask if
    [tex]
    \sqrt{4} = -\sqrt{4}
    [/tex]
    and, if [tex]\sqrt{4} = \pm 2[/tex],
    [tex]
    +(\pm 2) = - (\pm 2)
    [/tex]

    which is a true statement. But this would mean -x = x with x nonzero, which is false. So this means [tex]\sqrt{4}[/tex] has only one value. Is this kinda sorta right?
     
  8. Apr 9, 2015 #7
    No, square root of 4 is 2 and only 2, but 2^2 OR (-2)^2 will be equal to 4.

    So if you're confronted with something like √x =2, then x is 4. The square root function cannot return a negative real number.

    On the other hand, if you've got x^2 = 4, then both x = 2 and x = -2 are valid solutions to that algebraic equation. If you were to solve that equation by taking the square root of both sides, you would have to include x = -2 along with the x = 2 that would be returned by that operation because, unlike the square root function, the squaring function has values for all real numbers.
     
  9. Apr 9, 2015 #8

    Mark44

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    A better way to solve the equation ##x^2 = 4## is to write it as ##x^2 - 4 = 0## or (x - 2)(x + 2) = 0, from which we get x = 2 or x = -2. No funny business with taking the square root of both sides needed.
     
  10. Apr 9, 2015 #9

    Mentallic

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    If [itex]\sqrt{4}=\pm 2[/itex], then when using the quadratic formula

    [tex]x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

    Why would we need the [itex]\pm[/itex] symbol there considering [itex]\sqrt{b^2-4ac}[/itex] should give us the positive and negative value?
     
  11. Apr 10, 2015 #10

    BobG

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    Maybe what's missing from this thread is the definition of a function.

    https://www.mathsisfun.com/definitions/function.html

    Accordingly, the arcos of 0.5 is 60 degrees because the range of the arcos function is 0 to 180 degrees. None the less, if you're looking at a circle, there are two angles (-60 and +60) that have a cosine of 0.5.
     
  12. Apr 10, 2015 #11
    Man I love this board... :) I'm guessing some of our spouses would have left us by now if we didn't have this forum and phys.org! :)
     
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