Animated Solution for Mathematics - Quadratic Equations: The sum of all the real roots of the equation $(e^{2x}-4)(6e^{2x}-5e^x+1)=0$ is

Imagine you are standing before a complex, intimidating equation:

Your first instinct might be to expand it, to multiply everything out and see what happens. But stop! In the world of JEE Advanced, expansion is often a trap.

Instead, we use the Zero Product Property. This elegant principle tells us that if the product of two expressions is zero, then at least one of those expressions must be zero. We have just turned one difficult problem into two manageable ones.

The First Branch

Let us tackle the first factor: $e^{2x}-4=0$. This is a classic exponential equation.

By moving the constant to the other side, we get $e^{2x}=4$. Taking the square root of both sides, we find $e^x=2$.

Since $e^x$ is always positive, we do not need to worry about negative roots. Taking the natural logarithm, we arrive at our first root:

The Quadratic Mask

Now, look at the second factor: $6e^{2x}-5e^x+1=0$. At first glance, it looks like a mess of exponentials.

But look closer—it is a quadratic equation in disguise! By using the substitution $t=e^x$, the equation transforms into:

This is a standard quadratic that we can solve by splitting the middle term. We factor it as $(3t-1)(2t-1)=0$, which gives us $t=\frac{1}{3}$ and $t=\frac{1}{2}$.

Substituting back $t=e^x$, we find our next two roots:

The Logarithmic Symphony

We have our three roots: $\ln 2$, $\ln (\frac{1}{3})$, and $\ln (\frac{1}{2})$. The question asks for their sum.

We could calculate them individually, but there is a more beautiful way. Using the logarithmic property $\ln a+\ln b+\ln c=\ln (abc)$, we can combine them into a single expression:

The arithmetic here is a delight—the $2$ in the numerator and the $2$ in the denominator cancel out, leaving us with $\ln (\frac{1}{3})$.

Finally, using the property $\ln (\frac{1}{a})=-\ln a$, we reach our final answer:

You see? By staying calm and using the right tools, even the most intimidating problems reveal their elegant, simple core. Keep practicing, and you will master this art.