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Why do we use plus/minus signs in front of radicals?

  1. Jan 24, 2013 #1
    [tex]±\sqrt { { x }^{ 2 } } \\ =±|x|\\ =±±x[/tex]

    Wouldn't a plus sign or a minus sign be sufficient?
    Last edited by a moderator: Jan 24, 2013
  2. jcsd
  3. Jan 24, 2013 #2


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    hi tahayassen! :smile:
    you mean, wouldn't one plus-or-minus sign be sufficient?

    yes, you can just write "√(x2) = ±x" :wink:
  4. Jan 24, 2013 #3
    The last step is either false or misleading nobody would write that. It means in one case ++x and in the other case --x which is both the same as x and not what you wrote on the line above. Plus-minus signs are used if one wants to say that there are two equations that apply in different cases, and that are only distinguished by different plus or minus signs. It's a way to express your thoughts and not a rigorous 100% correct way to write mathematics. If one wants to be clear one has to write cases like

    "c=a+b if condiction X and c=a-b if condition Y."
  5. Jan 24, 2013 #4
    You should also note that plus/minus signs written as in your thread title means something different.

    +/- means plus or minus, but not both, ie only one will satisfy the statement.

    So an object that slides up and down a vertical pole can be moving up (y +ve) or down (y -ve) but not both.

    ± means both plus and minus (satisfy the statement)

    The equation x2 = 4 is satisfied by both x=+2 and x=-2
  6. Jan 24, 2013 #5
    Plus/minus signs are not aways used in front of radicals. The n-th root is a function which means it can produce only one output. In the case where n=even number the square root is positive number which rised to the power of n gives the original number. However negative numbers risen to even power give the same answer as the same positive numbers. When you solve equations you usualy need both cases.

    And yes +/- sign mean both cases separately for example :

    [itex](x \pm y)^2=x^2 \pm 2xy+ y^2[/itex] This is basicaly the two cases in one equation :
    [itex](x + y)^2=x^2 + 2xy+ y^2[/itex]
    [itex](x - y)^2=x^2 - 2xy+ y^2[/itex]
  7. Jan 25, 2013 #6


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    This last step is incorrect. [itex]\pm\sqrt{x^2}[/itex] is the same as [itex]\pm |x|[/itex] but |x| is NOT the same as "[itex]\pm x[/itex]" so [itex]\pm |x|[/itex] is NOT the same as "[itex]\pm\pm x[/itex]".

    How would you chose which?
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