Real Solution of x^3+8 == 0?
I do not understand the following:
sage: assume(x,'real')
sage: solve(x^3+8==0,x)
[]
Why does this equation have no solution? But -2 is a solution!
Thanks for help!
I do not understand the following:
sage: assume(x,'real')
sage: solve(x^3+8==0,x)
[]
Why does this equation have no solution? But -2 is a solution!
Thanks for help!
I think this is because (-1)^(1/3)
is not considered to be real.
sage: solve(x^3+1==0,x)
[x == 1/2*I*(-1)^(1/3)*sqrt(3) - 1/2*(-1)^(1/3), x == -1/2*I*(-1)^(1/3)*sqrt(3) - 1/2*(-1)^(1/3), x == (-1)^(1/3)]
sage: assume(x,'real')
sage: solve(x^3+1==0,x)
[]
Note that Maxima (which does our solving) doesn't actually care about x
being real, since it's a dummy variable.
(%i1) declare(x,real);
(%o1) done
(%i2) solve(x^3+1=0,x);
sqrt(3) %i - 1 sqrt(3) %i + 1
(%o2) [x = - --------------, x = --------------, x = - 1]
2 2
But when it's returned to Sage, somehow it doesn't keeps the x=-1
syntax and gets the cube root again, and it falls prey to
sage: (-1)^(1/3).n()
0.500000000000000 + 0.866025403784439*I
This is now http://trac.sagemath.org/sage_trac/ticket/11941.
An alternative approach is not to use the symbolic ring but a polynomial ring:
sage: x = polygen(RR)
sage: (x^3+8).roots()
[(-2.00000000000000, 1)]
This returns a list of roots in RR with multiplcities.
Please start posting anonymously - your entry will be published after you log in or create a new account.
Asked: 2011-10-18 15:07:48 +0100
Seen: 2,857 times
Last updated: Dec 11 '13
exponential equation real solution
Issues with: Solving a polynomial equation with multiple variables
How to get all (numerical) solutions of an equation?
Using the solution of equation
Solve system of equations with additional conditions in sage
wxMaxima cannot reduce system to a polynomial in one variable [closed]
How to make solve to use certain variables on the right side