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Square rooting an equation/inequality, which side is +/-?

  1. Apr 12, 2012 #1
    When you have something like x^2 > 2/3 and you root it, why does the left hand side become the plus or minus and not the right side? I only get the correct answer if I do it on the left hand side.
     
  2. jcsd
  3. Apr 12, 2012 #2

    micromass

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    I think the usual way of doing this is not writing +/- in front of anything, but just expanding this equations in two equations.

    Let's take an x such that

    [tex]x^2>4[/tex]

    (I pick 4 because the notation will be easier). When I square root both sides, then I'll get

    [tex]\sqrt{x^2}>2[/tex]

    However, note the very important point that [itex]\sqrt{x^2}[/itex] is NOT always equal to x. This is only true if [itex]x\geq 0[/itex]. For example: [itex]\sqrt{(-1)^2}=1\neq -1[/itex].

    So we have to split up in two cases:
    1) Either [itex]x\geq 0[/itex]. In that case [itex]\sqrt{x^2}=x[/itex], so we get

    [tex]x>2[/tex]

    2) Or x<0, in that case [itex]\sqrt{x^2}=-x[/itex], so we get

    [tex]-x>2[/tex]

    Or equivalently

    [tex]x<-2[/tex]

    So, to conclude: if we take an x such that [itex]x^2>4[/itex], then we either get that x>2 or x<-2.
     
  4. Apr 13, 2012 #3

    Mark44

    Staff: Mentor

    To expand slightly on what micromass said, while it is not true that
    ##\sqrt{x^2} = x##
    it is true that ##\sqrt{x^2} = |x|##.

    Getting rid of the absolute values leads to the same two cases that micromass mentioned.
     
  5. Apr 16, 2012 #4
    ^ Thanks, but don't understand why you can't put a plus/minus (besides from the fact that you get an incorrect answer :p) in front of the number
     
  6. Apr 16, 2012 #5

    Mark44

    Staff: Mentor

    Because, by definition, the square root of a nonnegative real number is nonnegative.

    For example, many people erroneously believe that √4 = ±2. Although 4 does have two square roots, the principal square root of 4 is 2.
     
  7. Apr 16, 2012 #6

    HallsofIvy

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    It doesn't. If [itex]x^2> 2/3[/itex] then either [itex]x<-\sqrt{2/3}[/itex] or [itex]x> \sqrt{2/3}[/itex].

    Conversely, if the problem were [itex]x^2< 2/3[/itex] then [itex]-\sqrt{2/3}< x< \sqrt{2/3}[/itex].
     
  8. Apr 23, 2012 #7
    ^ Thanks HallsOfIvy, never knew that!
     
  9. Apr 23, 2012 #8


    Well, it is NOT erroneous to "believe" that [itex]\sqrt{4}=\pm 2[/itex] since, as it happens, both values on the RHS when

    squared equal 4 and this is the primary definition of "square root.

    It is DEFINED that [itex]\sqrt{4}=2[/itex] mostly, I think, to make [itex]\sqrt{x}[/itex] a function, which otherwise it wouldn't be. If one want to mess with the

    negative root is thus customary to take [itex]-\sqrt{4}=-2[/itex] and everybody happy.

    DonAntonio
     
  10. Apr 23, 2012 #9

    Mark44

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    Granted, 4 has two square roots, and I mentioned this earlier in the thread. However, the notation ##\sqrt{a}## indicates the principal square root.
     
  11. Apr 28, 2012 #10
    Ooh, one more thing, does this apply to trig equations as well?

    e.g. cos^2(x) = 1/4 becomes cos(x)=1/2 and -cos(x)=1/2 (from ±√(cos^2(x)) = √(1/2)) rather than cos(x)=±1/2 (which evaluates to the same answer, just being pedantic)
     
  12. Apr 29, 2012 #11

    Mark44

    Staff: Mentor

    Yes, of course. The above should be cos(x) = 1/2 OR -cos(x) = 1/2.
     
  13. Apr 29, 2012 #12
    ^ Thank you! Ah, missed that >.<
     
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