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The Square Root of Numbers

  1. Dec 8, 2011 #1
    1. The problem statement, all variables and given/known data

    What I am confused about is:
    [itex] x - \sqrt{2}=0[/itex]
    [itex]x=\sqrt{2}[/itex]
    is it equal to
    [itex]x=\pm\sqrt{2}[/itex]

    ????
    2. Relevant equations



    3. The attempt at a solution

    The way I know it, those two are not equal. But my teacher insists on saying that whenever there is a square root, we should always recognize also the negative root.

    It says based from this post : https://www.physicsforums.com/showpost.php?p=3618911&postcount=2 that it isn't.

    Isn't it you only consider to get the positive and negative root if it's like this:
    [itex]x^2=25[/itex]
    Then that's when you say that [itex]x=\pm5[/itex]

    Is there any law or postulate that clears this situation??
    Please correct me if I am wrong and if my teacher is actually correct.

    And if I am correct, how can I explain it to her so that she'll get it right away? I'm not that good at explaining anyway.
     
  2. jcsd
  3. Dec 8, 2011 #2

    how is: [itex]x=\pm5[/itex]

    different from: [itex]x=\pm \sqrt{25}[/itex]
    ?
     
  4. Dec 8, 2011 #3
    No, what I meant in my post is when an equation is not quadratic, as :

    [itex]x-\sqrt{25}=0[/itex]

    is different from when the equation is quadratic:

    [itex]x^2-25=0[/itex]

    So, the solution for the quadratic equation is:
    [itex]x=\pm5[/itex]

    I'm asking if the solution for the first equation will be:
    [itex]x=\pm\sqrt{25}[/itex]
    or is it just the positive one??

    If so, what is the explanation?
     
  5. Dec 8, 2011 #4
    I am perhaps not the best to answer this, but in my opinion(which doesn't count alot in math) It could be either, because -5^2 = 25, so i we use -5, x+5 =0, x = -5, if we +5 we have x-5 = 0, x = 5

    You could just write [itex] x = \sqrt{25} = \pm 5[/itex]
     
  6. Dec 8, 2011 #5

    I like Serena

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    Hi frozonecom! :smile:

    Obviously
    [itex]x=\sqrt{2}[/itex]
    is not the same as
    [itex]x=\pm\sqrt{2}[/itex]


    But if you solve
    [itex]x^2=25[/itex]
    the solution is
    [itex]x=\pm\sqrt{25}[/itex]


    Note that the square root function of a positive real number is defined to be always positive.
     
  7. Dec 8, 2011 #6

    dextercioby

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    Can it be the minus ? Think. What does the solution of an equation means ? If you plug the solution back to the equation, you should get something like 0=0, or other trivial equality.

    So x-√25=0. How do you solve this equation ? What is the solution ?
     
  8. Dec 8, 2011 #7

    Mark44

    Staff: Mentor

    No.
    If [itex] x - \sqrt{2}=0[/itex], then [itex] x = \sqrt{2}[/itex]. Period.
    x is a positive number, approximately 1.414.
     
  9. Dec 8, 2011 #8

    Ray Vickson

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    You need to distinguish between "root of a number" and the "square-root *function*". Every piece of software I know of treats the square-root _function_ (sqrt or sqr or ....) as a single POSITIVE value (provided that you start with a positive value whose root you want). However, there are two *roots* to the equation x^2 = a (a > 0), namely: x = sqrt(a) and x = -sqrt(a). '

    RGV
     
  10. Dec 8, 2011 #9
    frozonecom,

    Only when you take the square root of something do you put ±. If there already is a square root, leave it as it is.
     
  11. Dec 8, 2011 #10

    eumyang

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    I'm no expert, but I do know that the bolded is not correct. -5^2 = -25. Do you see why?
     
  12. Dec 8, 2011 #11

    Mark44

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    eumyang is correct. -52 means - (52) = -25. On the other hand, if you mean the square of -5, write it as (-5)2.
    No, this is WRONG. √25 = 5. Period.

    It's true that 25 has two square roots, but the symbol √25 means the principal (or positive) square root. The other square root is written -√25.
     
  13. Dec 9, 2011 #12
    Problem solved. Thank you for all your time in explaining.

    I guess I was just a little confused.
     
  14. Dec 9, 2011 #13
    But if we have [itex]\sqrt{x} ; x = 25[/itex] Then the answer is [itex]\pm 5[/itex]? So that it depends on whether we are taking the root of a number or variable? Or is it only if we have a value squared, and we put the squareroot sign that we use [itex]\pm[/itex]?
     
  15. Dec 9, 2011 #14

    gb7nash

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    No. The definition of √x (x > 0) is the positive number y such that y2 = x.

    However, if you're given something such as x2 = 100, there exist two solutions. √100, which is +10, and -√100, which is -10
     
    Last edited: Dec 9, 2011
  16. Dec 9, 2011 #15

    NascentOxygen

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    This is a good question, frozonecom. Apparently there are regional differences, so the correct answer depends on the country you are in (or the country where your teacher studied).

    The answers you received in this forum indicate the square root symbol to be taken as denoting the positive square root. I, however, studied along with the same nomenclature as your teacher, and I require recognition of both + and - answers for the explicit square root symbol.

    Another area where there are regional differences is in expressions such as 2/5x
    On some forums this is staunchly defended as being equal to 0.4/x
    On this forum, I'm pleased to see it is just as strongly regarded to be 0.4x

    With clarity being crucial to mathematics, I endorse only the latter.
     
  17. Dec 9, 2011 #16

    cepheid

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    If clarity is crucial to mathematics, then it would seem that you ought to endorse never typing 2/5x, and always only typing either 2/(5x) or (2/5)x.
     
  18. Dec 9, 2011 #17

    cepheid

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    Regarding the original post, if [itex] x - \sqrt{2} = 0 [/itex], then [itex] x = \sqrt{2} [/itex], and that's IT. Surely this is true REGARDLESS of whether you conventionally think of [itex] \sqrt{2} [/itex] as denoting only +1.414..., or whether you think of it as something that could equally well denote either +1.414.. or -1.414...

    Either way, the answer is [itex] x = \sqrt{2} [/itex]. An answer of [itex] -\sqrt{2} [/itex] wouldn't make sense, because as dextercioby has already pointed out, plugging this value back into your original equation, you see that it is not a solution:[tex] -\sqrt{2} - \sqrt{2} \neq 0 [/tex]

    Now, for those of you who apparently adopt the convention that the radical sign can denote either the positive or the negative root, then I suppose you could argue that the first radical is meant to denote -1.414..., and the second radical is meant to denote +1.414..., in which case you do get 0. However, in that case, you have the same symbol used to represent two different numbers in the same equation, and nothing whatsoever to indicate which one is which. This seems like a really bad notational practice to me.

    The convention in which the square root sign denotes only the positive root makes a lot more sense to me.
     
  19. Dec 10, 2011 #18
    Do we do the same thing when it is a trigonometric equation??? if:

    sin(x) - sqrt(3)/2 = 0

    then the solution will be the angles where sin is positive which is on QI and QII.
     
  20. Dec 10, 2011 #19

    eumyang

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    Questions like this also need statement regarding restrictions for x (like -π/2 ≤ x ≤ π/2 or x in ℝ). If there was a restriction of -π/2 ≤ x ≤ π/2, for instance, then you shouldn't look for the angle in Q II at all.
     
  21. Dec 11, 2011 #20

    NascentOxygen

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    While you are a student, you calculate answers the way your lecturer wants you to. :eek::)

    If you are unsure, then I think in an exam it is worth playing safe. Work it out assuming the radical means the positive only. Then, write, "if we are required to also consider the negative of the square root then ...." and provide a further solution.

    You are likely to lose few marks if your lecturer doesn't entertain the negative alternative, and you should get full marks if your lecturer expects that you'll examine both outcomes.
     
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