If 2a + 3b + 6c = 0,then at least one root of | vedclass

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(A) Let $f(x) = ax^2 + bx + c$.Define $F(x) = \int_{0}^{x} f(t) dt = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx$.Clearly,$F(0) = 0$.Also,$F(1) = \frac{a}{3}

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$(0, 1)$

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(A) Let $f(x) = ax^2 + bx + c$.Define $F(x) = \int_{0}^{x} f(t) dt = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx$.Clearly,$F(0) = 0$.Also,$F(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a + 3b + 6c}{6}$.Given $2a + 3b + 6c = 0$,we have $F(1) = 0$.Since $F(0) = F(1) = 0$ and $F(x)$ is a polynomial (and thus continuous and differentiable),by Rolle's Theorem,there exists at least one $x \in (0, 1)$ such that $F'(x) = 0$.Since $F'(x) = f(x) = ax^2 + bx + c$,there exists at least one root of $ax^2 + bx + c = 0$ in the interval $(0, 1)$.

If $2a + 3b + 6c = 0$,then at least one root of the equation $ax^2 + bx + c = 0$ lies in the interval:

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