Imaginary i

  • #1
TKDKicker89
2
0
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.
sq.rt.(-x^2-4x-3)
 

Answers and Replies

  • #2
yourdadonapogostick
270
1
i would start by factoring out the -1 and seeing if i can't factor the polynomial more.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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TKDKicker89 said:
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.
sq.rt.(-x^2-4x-3)

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.
 
  • #4
yourdadonapogostick
270
1
[tex]\sqrt{1-(x+2)^2}[/tex] can be simplified more
 
  • #5
HallsofIvy
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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]
 
  • #6
Delta
87
0
or even [tex]\sqrt{(-1-x)(3+x)}[/tex] :wink:
 

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