If a,b,c,∈R and the equation x2+(a+b)x+c=0 has no real roots, then
c(a+b+c) then more
2
-2
0
none of these
We have f(x)=x2+(a+b)x+c>0∀x∈R
Thus, f(0)>0,f(1)>0
⇒c(a+b+c)>0