If 2a + 3b + 6c = 0 and a, b, c in mathbb{R}, | vedclass

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(C) Given the condition $2a + 3b + 6c = 0$.Dividing by $6$,we get $\frac{a}{3} + \frac{b}{2} + c = 0$.Let $f(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx

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$ax^2 + bx + c = 0$

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(C) Given the condition $2a + 3b + 6c = 0$.Dividing by $6$,we get $\frac{a}{3} + \frac{b}{2} + c = 0$.Let $f(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx$. Note that $f(x)$ is a polynomial function,so it is continuous on $[0, 1]$ and differentiable on $(0, 1)$.Calculate the values at the endpoints:$f(0) = 0$.$f(1) = \frac{a}{3} + \frac{b}{2} + c = 0$.Since $f(0) = f(1) = 0$,by Rolle's Theorem,there exists at least one $c_0 \in (0, 1)$ such that $f'(c_0) = 0$.$f'(x) = ax^2 + bx + c$.Therefore,the equation $ax^2 + bx + c = 0$ has at least one root in $(0, 1)$.

If $2a + 3b + 6c = 0$ and $a, b, c \in \mathbb{R}$,then the equation $ax^2 + bx + c = 0$ has at least one root between $0$ and $1$.

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