Animated Solution for Mathematics - Quadratic Equations: 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:

Imagine you are standing before a parabola, a graceful curve that defines the quadratic polynomial $f(x)$. In the world of JEE Advanced, we often get lost in the sea of coefficients $a$, $b$, and $c$, but today, we are going to look at the problem through the lens of its roots.

The problem gives us a gift: one root is $-1$. We do not know the other, so we call it $k$.

Instead of the standard form, we embrace the factored form:

This is not just an equation; it is a map. It tells us exactly where the parabola kisses the $x$-axis.

The problem provides a condition: $f(-2)+f(3)=0$. This is a beautiful piece of information.

Geometrically, it tells us that the height of the parabola at $x=-2$ and the height at $x=3$ are equal in magnitude but opposite in direction. One is above the axis, one is below, and they perfectly balance each other out.

Let us perform the substitution:

For $x=-2$:

For $x=3$:

Now, we combine them:

Here is where the magic happens. We see an $A$ in every term. Since $f(x)$ is a quadratic, $A$ cannot be zero.

We divide by $A$, and just like that, the complexity vanishes. We are left with a simple, linear equation:

Expanding this, we get:

Combining like terms, we have $-3k+14=0$, which gives us $3k=14$, or:

We have found our mystery root! The roots are $-1$ and $\frac{14}{3}$.

The question asks for the sum of the roots. We add them:

We have arrived at our destination. The journey was not about brute force; it was about choosing the right tool—the factored form—and trusting the algebra to reveal the hidden structure.

Keep this in mind for your next exam: when you see roots, think factored form. It is the shortest path to the truth. The final answer is $\frac{11}{3}$.