Given:
#alpha+sqrtbeta# is a root of #x^2+px+q=0#
#alpha-sqrtbeta# is a root of #x^2+px+q=0#
we know hat
#-p/2+sqrt((p/2)^2-q)# is a root of #x^2+px+q=0#
#-p/2-sqrt((p/2)^2-q)# is a root of #x^2+px+q=0#
Comparing
#alpha=-p/2#
#beta=(p/2)^2-q#
If #1/(alpha+sqrtbeta)# and #1/(alpha-sqrtbeta)# are the roots
Then
#x-1/(alpha+sqrtbeta)# is a factor
#x-1/(alpha-sqrtbeta)# is a factor
Thus, the quadratic equation in the form of factors is
#(x-(1/(alpha)+1/sqrtbeta))(x-(1/(alpha)-1/sqrtbeta))=0#
#x^2-((1/(alpha)+1/sqrtbeta)+(1/(alpha)-1/sqrtbeta))x+(1/(alpha)+1/sqrtbeta)(1/(alpha)-1/sqrtbeta)=0#
#x^2-(1/(alpha)+1/sqrtbeta+1/(alpha)-1/sqrtbeta)x+((1/(alpha))^2-(1/sqrtbeta)^2)=0#
#x^2-2/alphax+1/alpha^2-1/beta=0#
#x^2-2/alphax+(beta-alpha^2)/(alpha^2beta)=0#
#x^2-2/alphax-(alpha^2-beta)/(alpha^2beta)=0#
Multiplying throughout by #alpha^2beta#,
#alpha^2betax^2-2alphabetax-(alpha^2-beta)=0#
Substituting for #alpha# and #beta#, we have
#alpha=-p/2#
#beta=(p/2)^2-q#
#alpha^2=p^2/4#
#-2alphabeta=-2(-p/2)(p^2/4-q)=p(p^2/4-q)#
#alpha^2-beta=p^2/4-(p^2/4-q)=q#
#(p^2/4(p^2/4-q))x^2+p(p^2/4-q)x-q=0#
#(p^2/4)((p^2-4q)/4)x^2+p((p^2-4q)/4)x-q=0#
Simplifying by multiplying throughout by 16
#p^2(p^2-4q)x^2+4p(p^2-4q)x-16q=0#
#(p^2-4q)(p^2x^2+4px)-16q=0#
