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

Unlocking the Exponential Equation

My dear student, welcome to the arena of JEE Advanced mathematics. Today, we are going to dissect an equation that, at first glance, might seem like a chaotic mess of exponentials.

You see $e^{4x}+e^{3x}-4e^{2x}+e^x+1=0$ and your instinct might be to panic. But I want you to take a deep breath. In mathematics, especially in competitive exams, complexity is often just a mask for hidden beauty. Let us peel back that mask together.

The Hidden Pattern

Look closely at the coefficients: $1,1,-4,1,1$. Do you see it? The symmetry is staring right at us.

This is not a random collection of terms; it is a reciprocal equation in disguise. In the world of polynomials and exponentials, whenever you see coefficients that mirror each other, you are holding a golden key.

The strategy here is to break the symmetry by dividing the entire equation by the middle term. Why the middle term? Because $e^{2x}$ is the pivot point. Dividing by $e^{2x}$ will balance the powers on both sides, creating a beautiful harmony of terms.

The Algebraic Alchemy

Let us perform the division. When we divide $e^{4x}+e^{3x}-4e^{2x}+e^x+1=0$ by $e^{2x}$, we get:

Now, look at what we have created. We can group the terms with similar powers together. We pair $e^{2x}$ with $e^{-2x}$, and $e^x$ with $e^{-x}$.

The equation transforms into:

This is the moment of transformation. We are moving from the realm of raw exponentials into the realm of algebraic substitution. Let us define a new variable, $u=e^x+e^{-x}$.

The Substitution

If $u=e^x+e^{-x}$, what happens when we square $u$? Let us calculate $u^2$:

Since $e^x\cdot e^{-x}=1$, this simplifies beautifully to $u^2=e^{2x}+e^{-2x}+2$. Therefore, $e^{2x}+e^{-2x}=u^2-2$.

Now, substitute this back into our grouped equation:

Simplifying this, we arrive at the quadratic equation: $u^2+u-6=0$. This is a familiar friend. We can factorize this as $(u+3)(u-2)=0$. This gives us two potential values for $u$: $u=-3$ or $u=2$.

The Constraint Trap

Here is where the JEE examiners test your maturity. Many students will stop here and try to solve for $x$ using both values. But wait! We must respect the domain of our substitution.

We defined $u=e^x+e^{-x}$. Recall the AM-GM inequality (Arithmetic Mean $\ge$ Geometric Mean). For any positive number $a$, $a+\frac{1}{a}\ge 2$. Since $e^x$ is always positive, $e^x+e^{-x}$ must be greater than or equal to $2$.

This means $u=-3$ is an impossible value. It is an extraneous root, a ghost in the machine. We must reject it. The only valid value is $u=2$.

The Final Victory

Now, we solve for $x$ using $u=2$:

This equation holds true only when $e^x=1$, which implies $x=0$. There are no other real values of $x$ that satisfy this.

Thus, we have found exactly one real root. The journey from a terrifying exponential equation to a simple linear solution is complete. Remember, my student, the math is not about memorizing steps; it is about recognizing patterns, respecting constraints, and trusting the logic. You have mastered this problem.