If 2 a+3 b+6 c=0(a, b, c R) then the quadratic equation a x^2+b x+c=0 has

Step-by-step solution

Let f(x)= a x^3 3 + b x^2 2 +c x f(0)=0 and f(1)= a 3 + b 2 +c= 2 a+3 b+6 c 6 =0 Also f(x) is continuous and differentiable in [0,1] and [0,1] . . So by Rolle's theorem, f^ (x)=0 . i.e. a x^2+b x+c=0 has at least one root in [0,1]