If $2 a + 3 b + 6 c = 0 ,$ then the equation $a x ^ { 2 } + b x + c=0$ has at least one real root in

We can solve this by applying Rolle's Theorem, which requires us to find a function whose derivative matches the given quadratic. Let the function be $f(x) = ax^3/3 + bx^2/2 + cx$, so its derivative is $f'(x) = ax^2 + bx + c$. We evaluate this function at the endpoints of the interval $(0, 1)$, giving $f(0) = 0$ and $f(1) = a/3 + b/2 + c$. To use the given condition $2a + 3b + 6c = 0$, we multiply our result for $f(1)$ by 6 to get $2a + 3b + 6c$, which equals 0. Since $f(0) = f(1) = 0$, Rolle's Theorem states that there must be at least one root of $f'(x)$ in the interval $(0, 1)$. Therefore, the equation has at least one real root in $(0, 1)$.

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