Question 130949
{{{w^4-12w^2-2=0}}}


Let {{{u=w^2}}}, so:


{{{u^2-12u-2=0}}}


Complete the square:
{{{u^2-12u=2}}}


{{{u^2-12u+36=38}}}


{{{(u-6)^2=38}}}


{{{u-6=sqrt(38)}}} or {{{u-6=-sqrt(38)}}}


{{{u[1]=6+sqrt(38)}}} or {{{u[2]=6-sqrt(38)}}}


But {{{u=w^2}}}, so {{{w=sqrt(u)}}} or {{{w=-sqrt(u)}}}


{{{w=sqrt(u[1])=sqrt(6+sqrt(38))}}}, or


{{{w=sqrt(u[2])=sqrt(6-sqrt(38))}}}, or


{{{w=-sqrt(u[1])=-sqrt(6+sqrt(38))}}}, or


{{{w=-sqrt(u[2])=-sqrt(6-sqrt(38))}}}


Anytime you have a 4th degree polynomial equation, you must expect to have exactly 4 roots.  They may be 4 real roots, 2 pairs of complex congujate roots, or 2 real roots and 1 pair of complex conjugate roots.  Pairs of real roots might be identical (multiplicity greater than 1) but there will ALWAYS be 4 roots when the highest degree term is 4.