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March 21st, 2008, 12:48 PM   #1
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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?
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March 21st, 2008, 03:03 PM   #2
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[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]
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March 26th, 2008, 03:39 AM   #3
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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?
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March 26th, 2008, 10:40 AM   #4
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[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]
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March 26th, 2008, 11:57 AM   #5
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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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