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Square root

  1. Mar 6, 2010 #1
    Consider square root of 4.

    Can square root of 4 be +2 or -2?

    I asked that to my math teacher so he said: NO, square root of 4 is +2!

    But I can't really understand why it cannot be -2 since -2 squared is also 4. One thing I've imagined is: square root of 4 can be +2 or -2, but for CONVENTION we use +2. Is that right?

    There's another question related to that: consider

    [tex]i^2 = -1[/tex]

    squaring the two sides we obtain

    [tex]i^4 = 1[/tex]


    [tex]i = 1[/tex]

    so: can the i (the imaginary number) be 1? I think I'm making a lot of confusion with that, could someone explain me that?

    Thank you
  2. jcsd
  3. Mar 6, 2010 #2
    The square root of 4 can be +2 or -2, your math teacher is retarded. What you did with the imaginary number results in what I believe is called an extraneous solution (correct me if I'm wrong). If you substitute the 1 back into your original problem i^2=-1, you will find that 1=-1, which is not a true statement, meaning the solution is incorrect.
  4. Mar 6, 2010 #3


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    Gold Member

    We get these types of questions a lot so it might be worthwhile to run a forum search on this topic to get a very thorough review of it. In fact, I might even say that this question has been discussed ad nauseum :wink:

    However, for your first question the, square root of a number is often defined as the principal (or positive) root. This convention is not without reason though; if we would like to have a square root function, the square roots must necessarily be single-valued. Therefore, [itex]\sqrt{4} = 2[/itex]. For a more complicated example, let [itex]x[/itex] be a number such that [itex]x^2 = 4[/itex], then [itex]\sqrt{x^2} = |x| = 2[/itex] because [itex]\sqrt{x^2}[/itex] is necessarily positive. Since [itex]|x| = 2[/itex], this means that [itex]x = 2[/itex] or [itex]x = -2[/itex].

    For your second question, squaring equations has the unfortunate effect of adding extraneous solutions. Taking your example, clearly [itex]i^4 - 1 = 0[/itex] in which case [itex](i^2 + 1)(i^2 - 1) = 0[/itex]. By the zero product property, either [itex]i^2 + 1 = 0[/itex] or [itex]i^2 - 1 = 0[/itex] (note that only one of these need hold). From the conventional definition of [itex]i[/itex] we have that [itex]i^2 + 1 = 0[/itex] and [itex]i^2 - 1 = 0[/itex] is just an extraneous solution added when we squared the equation.

    Hopefully this all makes sense.
  5. Mar 6, 2010 #4
    Thank you for the help guys. Now things make more sense.

    PS: next time i'll do a forum search...
  6. Mar 7, 2010 #5


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    If his teacher is retarded, what does that make you? His teacher is completely right. The equation [itex]x^2= 4[/itex] has two roots. But only one of them is [itex]\sqrt{4}[/itex] because [itex]\sqrt{a}[/itex] is defined as "the positive root of the equation [itex]x^2= a[/itex]".

    The reason we have to write the solutions to that equation as "[itex]x= \pm\sqrt{a}[/itex] is that [itex]\sqrt{a}[/itex] alone does NOT mean both solutions. If it did we would not need the "[itex]\pm[/itex]".
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