If $(2a+3b+6c=0)$, $a, b, c \in R$, then quadratic equation $ax^{2}+bx+c=0$ has
- At least one root in $[0, 1]$
- At least one root in $(0, 1)$
- At least one root in $ [2, 3]$
- At least one root in $ [4, 5]$
Let f(x) = ax^2 + bx + c. Given 2a + 3b + 6c = 0, divide by 6 to get a/3 + b/2 + c = 0. This is equivalent to evaluating the integral of the quadratic or testing values. Specifically, f(0) = c and f(1) = a + b + c. Using the relation, one can show there is a root in (0, 1) by Rolle's theorem or intermediate value theorem on the antiderivative.
Let f(x) = a x^2 + b x + c, so f(0) = c and f(1) = a + b + c. The given condition is 2a + 3b + 6c = 0, which means 2(a + b + c) + b + 4c = 0, so 2f(1) + b + 4c = 0. We also know f(1/2) = a/4 + b/2 + c, and 12f(1/2) = 3a + 6b + 12c = 3(2a + 3b + 6c) = 0, meaning f(1/2) = 0. Because f(1/2) = 0, the quadratic equation has at least one root in (0, 1).