+cx Being polynomial, it is continuous and differentiable, also, f(0)=0 and f(1)=
a
3
+
b
2
+c ⇒f(1)=
2a+3b+6c
6
=0 (given) ∴f(0)=f(1) ∴f(x) satisfies all conditions of Rolle's theorem therefore f′(x)=0 has a root in (0,1) i.e. ax2+bx+c=0 has at lease one root in (0,1)