Thanks very much, Chris. That solves the problem.

On Fri, Jan 16, 2015 at 2:09 PM, Chris Smith <[email protected]> wrote:

> (actually, you don't need to factor, but doing so shows the quadratic
> nature of the thing). I get for approx. solutions, +/- 2.6.
>
> On Friday, January 16, 2015 at 1:06:18 PM UTC-6, Chris Smith wrote:
>>
>> The expression is quadratic in C0 only if C0 is positive. Try replace C0
>> with var('x', positive=True); factor the result, *then* solve for your two
>> values.
>>
>> On Wednesday, January 14, 2015 at 4:44:28 PM UTC-6, Junwei Huang wrote:
>>>
>>> Hi all, I have a symbolic expression as in polyC
>>>
>>> In [46]: polyC
>>> Out[46]: -(sqrt(2035953389603699)*sqrt(-1/C0**12)/2000000 -
>>> 4849598923/(54000000*C0**6))**(1/3) - 1 + 3599/(300*C0**2) -
>>> 368417/(18000*C0**4*(sqrt(2035953389603699)*sqrt(-1/C0**12)/2000000 -
>>> 4849598923/(54000000*C0**6))**(1/3))
>>>
>>> In [47]: N(polyC,2)
>>> Out[47]: -(23.0*(-1/C0**12)**0.5 - 90.0/C0**6)**0.33 - 1.0 + 12.0/C0**2
>>> - 20.0*(23.0*(-1/C0**12)**0.5 - 90.0/C0**6)**(-0.33)/C0**4
>>>
>>> In [48]: polyC.simplify()
>>> Out[48]: -1 - (54*sqrt(2035953389603699)*sqrt(-1/C0**12) -
>>> 9699197846/C0**6)**(2/3)/(600*(27*sqrt(2035953389603699)*sqrt(-1/C0**12)
>>> - 4849598923/C0**6)**(1/3)) + 3599/(300*C0**2) -
>>> 368417*2**(1/3)/(60*C0**4*(27*sqrt(2035953389603699)*sqrt(-1/C0**12) -
>>> 4849598923/C0**6)**(1/3))
>>>
>>> In [49]: solve(polyC.simplify(),C0)
>>> Out[49]: []
>>>
>>> In [50]: solve_poly_system([polyC.simplify()],C0)
>>> PolynomialError: C0**(-2) contains an element of the generators set
>>>
>>>
>>> But if you exam Out[46], polyC seems to be something like
>>> C0**2*(expr)-1. then C0 can be solved with two roots. Anything I missed?
>>> Thank you.
>>>
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