If 4a + 2b + c = 0, then the equation 3ax2 + 2bx + c = 0 has at least one real root lying in the interval

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(0, 2) Let f\left( x ight) = a x^3 + b x^2 + cx + d . . . . . \left( 1 ight) f\left( 0 ight) = d f\left( 2 ight) = 8a + 4b + 2c + d = 2\left( 4a + 2b + c ight) + d = d \left( \because \left( 4a + 2b + c ight) = 0 ight) f is continuous in the closed interval [0, 2] and f is derivable in the open interval (0, 2). Also, f(0) = f(2) By Rolle's Theorem, f'\left( \alpha ight) = 0 	ext{ for } 0 < \alpha < 2 	ext { Now }, f'\left( x ight) = 3a x^2 + 2bx + c \Rightarrow f'\left( \alpha ight) = 3a \alpha^2 + 2b\alpha + c = 0 	ext { Equation } \left( 1 ight) 	ext{has atleast one root in the interval } \left( 0, 2 ight) . 	ext { Thus,} f'\left( x ight) 	ext{must have root in the interval } \left( 0, 2 ight) .

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