Prev: Question about maximality principles, lattices, AC and its equivalents
Next: Geometrical-probability theory #4.23 & #239 Correcting Math & Atom Totality
From: Kent Holing on 26 Jul 2010 23:03
This lead me to Q(-(a+b))= 0 and R(-(a+b)) = +-1.
Here Q(x) = x^4 + a x^3 + b x^2 + c x + d = 0 and R(x) = 0 its cubic (Lagrange) resolvent.
Q(x) = (x-r)R(x) for r = -(a+b).
From: Kent Holing on 27 Jul 2010 06:30
It can be proved that if the cubic tem og Q(x) is missing and R(t) is irreducible, then Q(x) = 0 and R(x) = 0 CANNOT have a common root.
From: Kent Holing on 28 Jul 2010 01:27
For {a,b,c,d}={-(2r^2+r+1)/r,(r^2+r+1)/r,(2r^3+r^2+r-1),-r^2(r^2 +r+1)} for r a real number the quartic
x^4 + a x^3 + b x^2 + c x + d = 0 has a common roots (in fact three common) roots with its resolvent.

This example was forwarded to me by Kurt Foster (private communication).

If we shall have monic equations in Z[x] then r = +-1,
and we get the quartics given by
{a,b,c,d}={-4,3,3,-3} and {a,b,c,d}={2,-1,-3,-1}.
First  |  Prev  | 
Pages: 1 2
Prev: Question about maximality principles, lattices, AC and its equivalents
Next: Geometrical-probability theory #4.23 & #239 Correcting Math & Atom Totality