My Math Forum Looks like a polynomial....but it isn't

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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^3-19x^(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 non-negative 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 (8u-27)(u+1)=0 ==> u = 27/8 or u = -1. I leave it up to you to re-substitute 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 u-values. You're taking cube roots, which should have three complex solutions.

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