Let $$f(x)$$ be a quadratic polynomial such that $$f(-2) + f(3) = 0$$. If one of the roots of $$f(x) = 0$$ is $$-1$$, then the sum of the roots of $$f(x) = 0$$ is equal to

Let $$f(x)$$ be a quadratic polynomial with one root equal to $$-1$$.

Write the general form:

Since $$-1$$ is a root, we can write:

$$f(x) = a(x + 1)(x - r)$$

where $$r$$ is the other root and $$a \neq 0$$.

Apply the condition $$f(-2) + f(3) = 0$$:

$$f(-2) = a(-2 + 1)(-2 - r) = a(-1)(-2 - r) = a(2 + r)$$

$$f(3) = a(3 + 1)(3 - r) = 4a(3 - r)$$

Setting their sum to zero:

$$a(2 + r) + 4a(3 - r) = 0$$

$$a[(2 + r) + 4(3 - r)] = 0$$

Since $$a \neq 0$$:

$$2 + r + 12 - 4r = 0$$

$$14 - 3r = 0$$

$$r = \frac{14}{3}$$

Find the sum of the roots:

$$\text{Sum of roots} = -1 + \frac{14}{3} = \frac{-3 + 14}{3} = \frac{11}{3}$$

Hence, the sum of the roots is $$\dfrac{11}{3}$$.

The correct answer is Option A.