Question 134785
Using the quadratic formula, the roots of {{{px^2+qx+r=0}}} are:


{{{x = (-q +- sqrt( q^2-4pr ))/(2p) }}} and these two roots differ by 4, so:


{{{((-q + sqrt( q^2-4pr ))/(2p))-( (-q - sqrt( q^2-4pr ))/(2p) )=4}}}


{{{2(sqrt( q^2-4pr ))/(2p)=4}}}


{{{sqrt( q^2-4pr )/p=4}}}


{{{sqrt( q^2-4pr )=4p}}}


{{{q^2-4pr=16p^2}}}


{{{q^2=16p^2+4pr}}}


{{{q^2=4p(4p+r)}}}


In general, if the roots differ by {{{d}}}, then {{{q^2=dp(dp+r)}}}