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Imaginary i

  1. Aug 4, 2005 #1
    I'm having some trouble figuring out how to simplify the following problem.
    I know that i= the sq root of -1, and that i^2=-1, but i'm not sure how to approach this problem.
  2. jcsd
  3. Aug 4, 2005 #2
    i would start by factoring out the -1 and seeing if i can't factor the polynomial more.
  4. Aug 4, 2005 #3


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    Exactly what is the problem? To simplify [tex]\sqrt{-x^2- 4x- 3)}[/tex]?

    Any time you have something like this, involving a square root,even if it doesn't involve i, think about completing the square.

    -x2- 4x- 3= -(x2+ 4x)- 3 and we can see that we need to add (4/2)2= 4 inside the parentheses to complete the square. This is -(x2+ 4x+ 4- 4)- 3= -(x2+ 4x+ 4)+ 1=
    -(x+2)2+ 1. The square root can be written as
    [tex]\sqrt{1-(x+2)^2}[/tex]. I don't see much more that can be done and I don't see that it has directly to do with i. Even though the original -x2- 4x- 3 has all "negatives", this can be positive. If x lies between -3 and -1, -x2-4x- 3 will be positive and the square root will be real.
  5. Aug 4, 2005 #4
    [tex]\sqrt{1-(x+2)^2}[/tex] can be simplified more
  6. Aug 4, 2005 #5


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    Well, yes, of course, how foolish of me! [tex]\sqrt{1-(x+2)^2}= \sqrt{(1-(x+2))(1+(x+2))}= \sqrt{(1-x)(3+x)}[/tex]
  7. Aug 4, 2005 #6
    or even [tex]\sqrt{(-1-x)(3+x)}[/tex] :wink:
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