Animated Solution for Mathematics - Quadratic Equations: The equation $e^{4x}+8e^{3x}+13e^{2x}-8e^x+1=0,x\in \mathbb{R}$ has:

Imagine you are standing before a massive, intimidating wall of exponents: $e^{4x}+8e^{3x}+13e^{2x}-8e^x+1=0$. It looks like a chaotic mess of terms, but in the world of JEE Advanced, chaos is often just order in disguise.

The first step to conquering this beast is to see past the exponents. Notice how the powers of $e$ are all multiples of $x$. This is our golden ticket.

By letting $t=e^x$, we transform this exponential nightmare into a clean, manageable polynomial. However, there is a vital constraint here: since $e^x$ is always positive for any real $x$, our new variable $t$ must be strictly greater than zero ($t>0$). Keep this in your pocket; it will be our filter for the final answers.

Substituting $t$ into our equation, we get:

Now, this is a quartic equation. While we could try to factor it using complex methods, look at the coefficients: $1,8,13,-8,1$. They have a beautiful, near-symmetric structure, which is the hallmark of a reciprocal equation.

To unlock this, we divide the entire equation by $t^2$. Since we know $t>0$, we are not dividing by zero, so this is perfectly legal. The equation becomes:

Now, let us group the terms that share coefficients:

This is where the magic happens. We introduce a second substitution: $z=t-\frac{1}{t}$.

If we square this, we get $z^2=t^2+\frac{1}{t^2}-2$, which implies $t^2+\frac{1}{t^2}=z^2+2$. Substituting this into our grouped equation, we get:

This simplifies to the elegant quadratic:

Factoring this is a breeze: $(z+3)(z+5)=0$. Thus, $z$ can be $-3$ or $-5$.

We are almost there! We have two cases for $t$.

Case 1: $t-\frac{1}{t}=-3$, which leads to $t^2+3t-1=0$. Using the quadratic formula, we find:

Since $t>0$, we reject the negative root and keep $t=\frac{\sqrt{13}-3}{2}$.

Case 2: $t-\frac{1}{t}=-5$, which leads to $t^2+5t-1=0$. Again, the quadratic formula gives:

We reject the negative root and keep $t=\frac{\sqrt{29}-5}{2}$.

Both valid $t$ values are between $0$ and $1$. Since $x=\ln (t)$, and the logarithm of a fraction between $0$ and $1$ is always negative, both solutions for $x$ are negative. We have successfully tamed the beast: the equation has two solutions, and both are negative.