Comment #6 on issue 4003 by smi...@gmail.com: Cannot return general quartic
solution
http://code.google.com/p/sympy/issues/detail?id=4003
For posterity, here are those 4 equations for the equation x**4 + a*x**3 +
b*x**2 + c*x + d = 0:
TH = Rational(1, 3)
g = b**2 - 3*a*c + 12*d
al = 2*b**3 - 9*a*b*c + 27*c**2 + 27*a**2*d - 72*b*d
be = sqrt(-4*g**3 + al**2)
a4 = 2**TH*g/(3*(al + be)**TH)
a5 = ((al + be)/54)**TH
a3 = a**2/4 - 2*b/3
a1 = a3 + a4 + a5
a2 = 2*a3 - a4 - a5 - (-a**3 + 4*a*b - 8*c)/(4*sqrt(a1))
r1 = -a/4 - (sqrt(a1) + sqrt(a2))/2
r2 = -a/4 + (-sqrt(a1) + sqrt(a2))/2
r3 = -a/4 + (sqrt(a1) - sqrt(a2))/2
r4 = -a/4 + (sqrt(a1) + sqrt(a2))/2
Consider this equation
eq=(x**4+a*x**3+b*x**2+c*x+d)
eq.subs(dict(a=2,b=3,c=z + 11,d=5))
x**4 + 2*x**3 + 3*x**2 + x*(z + 11) + 5
There is no real value of z that will give this more than 2 real roots and
yet all 4 roots given by the above formula for z=0, for example, are real:
zans=solve(_,x)
[w.subs(z,0).n(2) for w in zans]
[-2.3, -0.51, -0.49, 1.3]
Again, consider
eq = x**4 - 8*x**3 + 24*x**2 - x*(z + 32) + 16
zans=solve(eq, x, check=0)
[w.subs(z,0).n(2) for w in zans]
[nan, nan, nan, nan]
versus
[w.n(2) for w in solve(eq.subs(z,0))]
[2.0]
[w.subs(z,1).n(2) for w in zans] # 4 imaginary
[1.8 - 1.2*I, 1.8 + 1.2*I, 2.2 - 1.2*I, 2.2 + 1.2*I]
versus
[w.n(2) for w in solve(eq.subs(z,1))] # two imaginary
[1.0, 1.8 - 1.2*I, 1.8 + 1.2*I, 3.4]
I think it would be very unwise to use the general solution when it is not
generally applicable. That being said, the current routine can be a little
more permissive by checking p.is_nonzero rather than looking for a
numerical value to be tested when p.is_zero is not true.
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