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 equal to

The given equation is $e^{4x}-e^{3x}-4e^{2x}-e^x+1=0$. At first glance, it appears to be a complex arrangement of exponentials. However, in the context of JEE Advanced, such structures often hide an underlying symmetry.

The First Transformation

To simplify the landscape, we introduce the substitution $t=e^x$. The equation transforms into a fourth-degree polynomial:

Since $t=e^x$ and the exponential function is strictly positive, we must maintain the constraint $t>0$ throughout our calculations.

The Reciprocal Revelation

Observe the coefficients: $1,-1,-4,-1,1$. This is a reciprocal equation. To exploit this symmetry, we divide the entire equation by $t^2$ (which is valid since $t>0$):

This simplifies to:

By grouping the symmetric terms, we obtain:

The Second Substitution

Let us define a new variable $\alpha =t+\frac{1}{t}$. Squaring this expression yields ${\alpha }^2=t^2+\frac{1}{t^2}+2$, which implies $t^2+\frac{1}{t^2}={\alpha }^2-2$.

Substituting these into our grouped equation, we get:

This simplifies to the quadratic equation:

Factoring this quadratic gives $(\alpha -3)(\alpha +2)=0$.

The AM-GM Trap

We have two potential values for $\alpha$: $3$ and $-2$. However, we must respect our domain constraint. By the AM-GM inequality, for any $t>0$:

Since $\alpha$ must be at least $2$, the root $\alpha =-2$ is extraneous and must be discarded. We are left with the unique solution $\alpha =3$.

The Final Victory

We now solve $t+\frac{1}{t}=3$, which rearranges to:

The discriminant of this quadratic is $D=(-3{)}^2-4(1)(1)=5$. Since $D>0$, there are two distinct real roots for $t$.

Because both the sum of the roots ($3$) and the product of the roots ($1$) are positive, both values of $t$ are positive. Since $x=\ln (t)$, each positive $t$ yields exactly one real value for $x$.

Thus, there are exactly 2 real roots for the original equation.