Animated Solution for Mathematics - Quadratic Equations: If f(x) is a quadratic expression such that $f(1)+f(2)=0$, and -1 is a root of $f(x)=0$, then the other root of $f(x)=0$ is :-

Imagine you are standing before a parabola, a beautiful, symmetric curve defined by a quadratic expression $f(x)$. In the world of JEE mathematics, we often encounter problems that seem to demand brute force, but there is almost always a more elegant, structural way to view them.

Today, we are going to explore one such problem. We are given that $f(x)$ is a quadratic expression, one of its roots is $-1$, and it satisfies the mysterious condition $f(1)+f(2)=0$. Our goal is to find the other root, which we will call $\beta$.

When dealing with roots, the standard form $f(x)=a{x}^2+bx+c$ is often a trap. It forces us to juggle three unknowns ($a,b,c$).

Instead, let us use the factored form:

Here, $\alpha$ and $\beta$ are the roots. We already know one root is $-1$, so let us set $\alpha =-1$. Our expression immediately simplifies to:

By making this choice, we have reduced our problem to finding just one unknown, $\beta$, because $a$ will eventually take care of itself.

The problem gives us a condition: $f(1)+f(2)=0$. This is not just an equation; it is a constraint on the geometry of our parabola.

Let us calculate these values. Substituting $x=1$ into our expression, we get:

Similarly, substituting $x=2$, we get:

Now, we combine these into our condition:

Look at that $a$! It is present in both terms. Since $f(x)$ is a quadratic, we know $a eq 0$.

This is the moment of truth where the complexity vanishes. We can divide the entire equation by $a$, leaving us with a simple linear equation:

Now, it is just a matter of careful algebra. Expanding the brackets, we get:

Grouping the constants and the $\beta$ terms, we have:

With a quick move, we find $5\beta =8$, which gives us:

And there it is! The other root is $\frac{8}{5}$.

This problem is a perfect example of why we should always look for the most efficient representation of a function. By choosing the factored form, we turned a potentially messy system of equations into a straightforward linear solve. Keep this in your toolkit—whenever you see roots, think of the factored form first. You have got this!