If px2+qx+r=0 cannot have real roots then prove that r(p+q+r)>0
if it cannot have real roots
b2-4ac<0
q2-4pr<0
we take a function
f(x)=px2+qx+r
f(1)=p+q+r
now we should know that f(x) is a parabola and
f(1)>0 if r>0
f(1)<0 if r<0
r.f(1)>0 because
positive*positive=positive
and negetive*negetive=positive
Moderation Team
![[Post New]](/templates/default/images/icon_minipost_new.gif){kind=icon}
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
21
