<PRE><B>Question 173792</B>
The roots of {{{x^2+px+q=0}}} are {{{alpha}}} and {{{4alpha}}} if and only if {{{(x-alpha)}}} and {{{(x-4alpha)}}} are factors of {{{x^2+px+q}}}.  So:


{{{(x-alpha)(x-4alpha)=x^2-(alpha+4alpha)+4alpha^2=x^2-5alpha*x+4alpha^2=x^2+px+q}}}


Hence:  {{{p=-5alpha}}} and {{{q=4alpha^2}}}


{{{4p^2=4(-5alpha)^2=4*25alpha^2=100alpha^2}}}


{{{25q=25*4alpha^2=100alpha^2}}}


QED</PRE>