Question 993998
{{{x^2+2px+p^2-p=0}}}


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


{{{x=(-2p+- 2*sqrt(p^2-(p^2-p)))/(2)}}}


{{{x=(-p+- sqrt(p^2-p^2+p))/1}}}


{{{highlight(x=-p+- sqrt(p))}}}
and p is given to be a real number {{{p>0}}}.  This means the square root expression does not disappear, and the plus&minus nature of the two roots insures that x is two unequal values.


{{{x=-p-sqrt(p)}}}  or  {{{x=-p+sqrt(p)}}}