radical problem

Yes, but you made it a matter of choice, and so the "usual" translation from exponents to radicals no longer holds.

It is also true that \(\displaystyle (-1)^{2/3} = -1/2-\sqrt{3}/2i\) and that \(\displaystyle (-1/2-\sqrt{3}/2i)^{3/2} = -1\)
 
Re: JeffM

I thought you had done the math for the other one and saw it wasnt applicable, because the domain changed. the answer you gave was the correct one, and from you explaination, I was able to explain why one worked and one didnt.
Thanks to your help, my homework grades were 20/20, 20/20, 19/20,18/20.
Dont beat yourself up, you followed my logic and had it make logical sense to me :)

Thanks all.

cschulz57919
 
The principal complex root is real if x is non-negative, so 20.032 is the answer. 19.968 is only correct if one defines
23_11afc84ebef3eb2c91647b7f64ab0455.png
when
3_97fdf90850f660f05349f4ad145b62dc.png
which can cause "paradoxes"
Xa/b must equal both (b root of x)a and (b rootof xa)
 
This has been one of the most frustrating threads.I thought that I knew why some could disagree on how these were calculated. But then then I was reviewing some lessons for a computer algebra system built on MathCad, a widely used CAS in engineering schools. Playing around I typed in this very problem expecting to get one real solution. To my surprise it gave me two, see the attachmentuntitled.GIF.
 
I think there is disagreement of CAS's on what exactly a rational power means. My TI-89 emulator also gives two solutions, but a program like mathematica will give 1. Some will graph x^(2/3) as being positive for x<0, others it will be undefined. I'm over it though.
 
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