if 2a+3b+6c=0(a,b,c are real),then the quadratic equation ax2+bx+c=0 has-

Consider F(x) = (a/3)x³ + (b/2)x² + cx

Now, F(1) = a/3 + b/2 + c = 1/6(2a + 3b + 6c) = 0 , by using the given condition.

F(0) = 0

Hence, Rolle's theorem is satified.

Thus, by Rolle's theorem, there exists at least one root in [0. 1] such that F'(x) = 0

Here, F'(x) = ax² + bx + c