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What are the solutions to x^4+9 = 0 ?

Date: 03/31/2003 at 20:21:07
From: Christina
Subject: What are the solutions to x^4+9=0 ?

What are the solutions to x^4+9 = 0 ?

My confusion has to do with how to work with complex numbers when the 
root is higher than two.


Date: 04/01/2003 at 09:21:56
From: Doctor Ian
Subject: Re: What are the solutions to x^4+9 = 0 ?

Hi Christina,

When I see 

  0 = x^4 + 9 

I see

  0 = (x^2)^2 - (3i)^2 

which is a difference of squares, which means I can do this:

  0 = (x^2 + 3i)(x^2 - 3i) 

The second factor is another difference of squares:

  0 = (x^2 + 3i)(x + sqrt(3i))(x - sqrt(3i))

And if you look at it the right way, the first factor is also a
difference of squares:

  0 = (x^2 - (-3i))(x + sqrt(3i))(x - sqrt(3i))
 
    = (x + sqrt(-3i))(x - sqrt(-3i))(x + sqrt(3i))(x - sqrt(3i))

So the four 4th roots of -9 would be

   sqrt(3i)      sqrt(-3i)

  -sqrt(-3i)    -sqrt(-3i)

Let's check to see if those work:

  [sqrt(3i)]^4  = (3i)^2  = -9

  [sqrt(-3i)]^4 = (-3i)^2 = -9
  
And I think you can see that the other two will work as well. 

So, what is sqrt(3i)?  Obviously, 
 
  sqrt(3i) = sqrt(3) * sqrt(i)

So what is sqrt(i)?  To find out, take a look at this answer from the 
Dr. Math archives:

   The Fourth Root of -1
   http://mathforum.org/library/drmath/view/53846.html 

Does this help? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Basic Algebra
High School Imaginary/Complex Numbers

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