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March 21st, 2008, 11:48 AM  #1 
Newbie Joined: Feb 2008 Posts: 8 Thanks: 0  Looks like a polynomial....but it isn't 8x^319x^(3/2)27=0 , and x is an element of the complex numbers Please someone help me with this question. If it's not a polynomial then what is it? How many roots does it have, and what are they? 
March 21st, 2008, 02:03 PM  #2 
Member Joined: Feb 2008 Posts: 89 Thanks: 0 
[color=darkblue]Hi: By definition, the terms of a polynmial are of the form cx^n, where c is a real number and n is a whole number (i.e., a nonnegative integer). Accordingly, 8x^3  19x^(3/2)  27 is not a polynomial. That said, we can solve the given equation nonetheless. Begin with the substitution u = x^(3/2). Since x^3 = [x^(3/2)]^2 = u^2, the equation becomes, 8u^2  19u  27 = 0, which factors as (8u27)(u+1)=0 ==> u = 27/8 or u = 1. I leave it up to you to resubstitute x^(3/2) into these equations for u, and solve for x. Regards, Rich B.[/color] rmath4u2@aol.com 
March 26th, 2008, 02:39 AM  #3 
Newbie Joined: Feb 2008 Posts: 8 Thanks: 0 
But how do you know when to stop? I don't know the number of roots it has, that's the problem. Because it's not a polynomial I can't say it only contains 3 roots in the complex world, it probably contains more, but how many more?

March 26th, 2008, 09:40 AM  #4 
Member Joined: Feb 2008 Posts: 89 Thanks: 0 
[color=darkblue]Hi Julien: Thanks for your input. So, do we have no way of knowing how many roots such a function has? Rich B.[/color] rmath4u2@aol.com 
March 26th, 2008, 10:57 AM  #5 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
I would expect 6 roots, three from each of the uvalues. You're taking cube roots, which should have three complex solutions.


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