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Solution to a simple equation

  1. Sep 1, 2012 #1
    1. The problem statement, all variables and given/known data

    X^2 + 1 = 0

    2. Relevant equations

    find the solution

    3. The attempt at a solution

    As simple as this:

    what is the solution to x^2 +1 = 0

    .... my question, why am I wrong in thinking x^2 = -1 as a possible solution?
  2. jcsd
  3. Sep 1, 2012 #2


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    X^2 = -1 is not a solution for "what is x", which is what is being asked for, it is a solution to "what is X^2"
  4. Sep 1, 2012 #3
    Oh I see.... Well, a solution to x^2 could be [tex]\pm 1[/tex]. So what is the solution, I am a bit confused. Would it be

    x = -1/2
  5. Sep 1, 2012 #4


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    Staff: Mentor

    (-1/2)² = 1/4, so that's not a solution to X² + 1 =0

    You are right in re-arranging your equation as X² = -1
    now take the square root of both sides.
  6. Sep 1, 2012 #5
    Some information is missing here. What is x supposed to be?? Is it rational? Real? Complex?
    Whether or not a solution exists to this problem depends on what you allow x to be.
  7. Sep 2, 2012 #6


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    No, it's not. You have already said that x^2= -1, not [itex]\pm 1[/itex]. Now you need to get x itself, not x^2. You need to "undo" the square- what's the opposite of squaring?

    You seem to be confused about what "solution to an equation" means. What level mathematics are you taking? Where did you get this problem? Have you studied complex numbers.
  8. Sep 3, 2012 #7
    So the solution is

    [tex]x = \sqrt{-1}[/tex]
  9. Sep 3, 2012 #8


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  10. Sep 3, 2012 #9


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    Homework Helper

  11. Sep 3, 2012 #10


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    Homework Helper

    To the OP: you know the symbol we use for [itex]\sqrt{-1}[/itex], right?
  12. Sep 3, 2012 #11
    Yes, it's the imaginary number. :)
  13. Sep 3, 2012 #12


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    Which imaginary number?

  14. Sep 4, 2012 #13


    Staff: Mentor

    In other words, what symbol do we use to represent ##\sqrt{-1}##?
  15. Sep 5, 2012 #14
    i = sqrt{-1}
  16. Sep 5, 2012 #15


    Staff: Mentor

  17. Sep 5, 2012 #16
    That's right! If x was supposed to be real then, the equation has no solution, but since we are considering complex numbers the equation has a solution!

    For the fundamental theorem of Algebra a polynomial form has always a solution in the set of the complex numebers.
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