if 2a+3b+6c=0(a,b,c are real),then the quadratic equation ax2+bx+c=0 has-
Asked by abhilipsa satpathy | 16th Jun, 2013, 03:20: PM
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
Answered by | 16th Jun, 2013, 10:31: PM
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