🚀 go-pugleaf

I have been working with quartic equations for some time,
and have a couple of questions that I thought some one may
know.  I have a set of equations that give the roots to a 
quartic equation completely.  When there are two real and
two imaginary roots, this set of equations can be used.  
Ofcourse, these equations come from the resolvent cubic.

When the equation in question has 4 real roots, or 4 
imaginary roots, an irreducible case arises, and a 
trigonometric solution is needed.

Now, lets look at 4 real roots.  The resolvent cubic has
3 real roots, and they are found by an expression that
includes the term cos(theta/3 + c*2/3*pi)  where c is 
either 0, 1, or 2.  
The three different c values yield three different 
angles, thus three different solutions for the cubic.  The
three solutions to the cubic are checked to see which 
solution provides positive terms in the original 4 
equations that give the roots of the quartic, and then
using that particular cubic solution, the quartic roots 
can be found.

My first QUESTION (which I am reasonably sure about) is,
in the 4 real roots case, is their exactly one resolvent
cubic root that works to yield the quartic roots, and is
it equal probability that c be 0, 1, or 2, for that cubic
solution.

The other case is when there are 4 imaginary roots of the
quartic.  My evidence seems to suggest the following:
 1. There are still three real roots of the resolvent cubic
 2. ONLY the first cubic solution, ie, the one you
    get when c = 0 above, yields positive values for
    the terms necessary in finding the quartic roots, at
    which time the final answer for each quartic root 
    has an imaginary part.
 
Number 2 above is the question I am really trying to resolve.

I think that this method of solving the quartic should be
familar to those who have worked with analytic solutions
of a quartic, so I have left out the exact equations,
for different forms and notation exist in different 
references.

Kelly Peterson