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Math Crisis!

  1. Nov 12, 2009 #1
    I have been trying for the past hour or so to figure out the solution to this equation:

    [tex] \sqrt{x} + 1 = 0[/tex]

    but to no avail! Here's what i tried...

    [tex] \sqrt{x} = -1 [/tex]

    [tex]\sqrt{x} = i^4 [/tex]

    [tex]x = \pm i^8 [/tex]

    [tex]x = \pm 1[/tex]

    But this is most obviously wrong. What did I do wrong and what is the real answer?
     
  2. jcsd
  3. Nov 12, 2009 #2
    I think you may be over thinking it, what is the inverse of a square root?
     
  4. Nov 12, 2009 #3
    If I square both sides from the beginning I still get x = +/- 1, neither of which solve the equation.
     
  5. Nov 12, 2009 #4

    Office_Shredder

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    Gold Member

    You seem to have stated that -1=i4. Check again
     
  6. Nov 12, 2009 #5
    Yea that was a typo, I've figured out what went wrong though. -1 just isn't in the range of the square root function. It has no solution.
     
  7. Nov 12, 2009 #6
    more importantly, x^2=y^2 does not imply x=y.
     
  8. Nov 12, 2009 #7
    [itex]x^2 = y^2 \Rightarrow x = \pm y [/itex]

    isn't if and only if?
     
  9. Nov 12, 2009 #8
    [tex]x^{2} = y^{2} \Rightarrow |x| = |y|[/tex]

    Isn't this correct though?


    Also, MATLAB gave the error 'Warning: Explicit solution could not be found.' when trying to solve the equation.
    And Microsoft Math gave x=1, which is clearly incorrect
     
    Last edited: Nov 12, 2009
  10. Nov 12, 2009 #9
    I said 'x^2=y^2 does not imply x=y'. What you are saying is that x^2=y^2 implies x=+-y. There is no contradiction here.



    in the original post, you had

    sqrt(x)=-1

    then you square both sides and got

    x=1 (another mistake in your post was the x=+-1 part; where did the -1 come from??)

    and from that you concluded that, if x=1, then

    sqrt(x)=-1

    which is false, because x^2=y^2 does not imply x=y.


    There was a post here a while ago, I can't find it, but it expressed a common misconception people make when solving equations. When you have something like x+4=8, and you subtract 4 from both sides to get x=4, you did not prove that

    'if x=4 then x+4=8' (1)

    you actually proved that

    'if x+4=8 then x=4' (2)


    And (1) is true only if the operation you peformed was injective (i.e. one to one). subtraction is injective, squaring both sides isn't.
     
  11. Nov 12, 2009 #10
    sqrt(x)= -1

    Square both side

    x=1.

    Then

    sqrt(1) = 1 or -1

    since

    1^2 = 1
    AND

    -1^2 = 1 (*)

    Using (*)

    sqrt (-1^2) = sqrt (1)

    ie -1 = sqrt(1)
     
  12. Nov 12, 2009 #11

    Mark44

    Staff: Mentor

    No. [itex]\sqrt{1} = 1[/itex]. By convention, the square root of a number is generally accepted to mean the principal, or positive square root.

    So sqrt(x) = -1 has no solution in the reals. If you look at the graph of y = sqrt(x), you'll see that the domain is x >= 0.
     
  13. Nov 13, 2009 #12

    Mentallic

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    Homework Helper

    Like Mark44 has said, only the principle root (i.e. the positive number) is taken. This means that there is no complex solution to this equation.
     
  14. Nov 13, 2009 #13
    I don't yet see why there can be no complex solution. Can this be proven rigorously?
     
  15. Nov 13, 2009 #14
    The right hand side of the equation is not in the domain of the left hand side. If you intend on extending the principal square root function to complex numbers, the problem becomes choosing which complex root is the principal root, as there is no "positive" or "negative" in regards to the complex plane.
     
  16. Nov 14, 2009 #15

    Mentallic

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    The only complex roots would be [itex]cis(2k\pi)[/itex] for all k integers, which in other words is 1. But since we've already defined by convention that the square root of a positive is only its positive root, then we cannot get the answer [itex]\sqrt{x}=-1[/itex]

    If it weren't for this limitation (which some authors don't uphold for convenience in their specific topics of mathematics) then we would have the only possible complex solution x=1
     
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