Animated Solution for Mathematics - Quadratic Equations: The number of real roots of the equation $e^{6x}-e^{4x}-2e^{3x}-12e^{2x}+e^x+1=0$ is :

Welcome, student. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of exponential functions.

You see an equation like

and your instinct might be to panic. But in the world of JEE Advanced, intimidation is just a mask. Behind that mask lies a beautiful, elegant structure waiting to be revealed.

The Hidden Symmetry

Let us look at the equation again:

The key to solving this is not to treat it as a random collection of terms, but to look for patterns. Specifically, look at the powers of $e^x$.

We have $6x,4x,3x,2x,x$. Notice the $e^{6x}$, the $-2e^{3x}$, and the $+1$. If we group these together, we get $(e^{6x}-2e^{3x}+1)$.

Does this look familiar? It is a perfect square! Specifically, it is $(e^{3x}-1{)}^2$. This is our first major breakthrough.

The Transformation

Now, what about the remaining terms? We have $-e^{4x}-12e^{2x}+e^x$.

Let us rearrange the original equation to group these:

Notice that $(e^{4x}-e^x)$ can be written as $e^x(e^{3x}-1)$. If we let $y=e^{3x}-1$, then our equation becomes:

Suddenly, the monster has been tamed. We have a simple quadratic equation in terms of $y$.

Factorization and Splitting

We need to factorize $y^2-e^{x}y-12e^{2x}=0$. We are looking for two numbers that multiply to $-12e^{2x}$ and add to $-e^x$.

Those numbers are $-4e^x$ and $3e^x$. So, the equation becomes:

This gives us two distinct cases: $y=4e^x$ and $y=-3e^x$. Substituting back $y=e^{3x}-1$, we get two new equations:

The Cubic Analysis

Let $t=e^x$. Since $e^x$ is always positive, we must have $t>0$. Our equations become:

Now, we analyze these cubics. For $h(t)=t^3-4t-1$, we find $h(2)=-1$ and $h(3)=14$. By the Intermediate Value Theorem, there is a root in $(2,3)$.

For $k(t)=t^3+3t-1$, we find $k(0)=-1$ and $k(1)=3$. There is a root in $(0,1)$. Since both cubics are strictly increasing for $t>0$, each provides exactly one positive root.

The Final Count

We have found two valid positive values for $t$. Since $t=e^x$, each positive $t$ gives exactly one real value for $x$ (because $x=\ln (t)$).

Therefore, our original equation has exactly two real roots.

You see? With patience and the right perspective, even the most intimidating problems yield to the beauty of mathematics. Keep practicing, and you will start seeing these patterns everywhere!