Hi,

I see some merit in the above discussion, but when I try the following 
sequence, I get a RootOf expression as in my original post.

In [1]: from sympy import *

In [2]: from sympy.abc import *

In [3]: print N(RootOf(x**5 + x**3 + 1, 1))
N(RootOf(x**5 + x**3 + 1, 1))


However, when I alter the import sequence, I am able to get an approximate 
answer. Cross checking WA <http://www.wolframalpha.com/input/?i=x5%2Bx3%2B1>, 
tells it is likely to correct.

In [3]: from sympy.abc import *

In [4]: from sympy import *

In [5]: print N(RootOf(x**5 + x**3 + 1, 1))
-0.217853219392291 - 1.16695124566485*I


Err, I do not understand sympy properly. ;)



On Wednesday, June 5, 2013 2:46:16 PM UTC+5:30, Abhishek kumawat wrote:
>
> When i solve  'x**5 + x**3 + 1', it shows solution as 
> [RootOf(x**5 + x**3 + 1, 0),
>  RootOf(x**5 + x**3 + 1, 1),
>  RootOf(x**5 + x**3 + 1, 2),
>  RootOf(x**5 + x**3 + 1, 3),
>  RootOf(x**5 + x**3 + 1, 4)]
>
> How to get solution of this equation using sympy.
> Image of the same is attached
>
>
> <https://lh6.googleusercontent.com/-X6chv6Uv09c/Ua8ByKhGqYI/AAAAAAAAAEo/wLGSedGo9jg/s1600/rootof.png>
>
>

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