Animated Solution for Mathematics - Quadratic Equations: Let $S$ be the set of all real roots of the equation, $3^x(3^x-1)+2|3^x-1|+|3^x-2|=0$. Then $S$:

Welcome, fellow explorer of the mathematical landscape. Today, we are standing before a formidable-looking equation:

At first glance, it might seem like a chaotic mess of exponentials and absolute values. In the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

The first thing that should catch your eye is the recurring presence of $3^x$. Whenever you see a repeated structure, think of it as a signal to simplify.

Let us define a new variable, $t=3^x$. Because $3^x$ is an exponential function, it is strictly positive for all real $x$. Therefore, our new variable must satisfy the constraint $t>0$.

With this substitution, our 'monster' equation transforms into a much friendlier polynomial form:

Think of the modulus function as a mirror that flips negative values to positive. The expressions inside, $t-1$ and $t-2$, are the 'critical points' where the mirror flips.

Specifically, $t-1$ changes sign at $t=1$, and $t-2$ changes sign at $t=2$. These points divide our number line into three distinct territories: $0<t<1$, $1\le t\le 2$, and $t>2$.

We must investigate each territory to see if any roots are hiding there.

In this region, both $t-1$ and $t-2$ are negative. The modulus bars force them to flip: $|t-1|=1-t$ and $|t-2|=2-t$.

Substituting these into our equation:

Expanding this, we find $t^2-t+2-2t+2-t=0$, which simplifies to $t^2-4t+4=0$. This is a perfect square: $(t-2{)}^2=0$, leading to $t=2$.

Since our condition for this act was $t<1$, this root is an imposter. We reject it.

Here, $t-1$ is non-negative, but $t-2$ is negative. Thus, $|t-1|=t-1$ and $|t-2|=2-t$.

Our equation becomes:

Expanding this, we get $t^2-t+2t-2+2-t=0$. The terms collapse beautifully, leaving us with $t^2=0$, or $t=0$.

Since $0$ is not in the range $[1,2]$, we reject this root as well.

In the land of $t>2$, both expressions are positive. The modulus bars vanish, leaving:

Expanding this, we get $t^2-t+2t-2+t-2=0$, which simplifies to $t^2+2t-4=0$. Using the quadratic formula:

Since $t$ must be positive, we take $t=\sqrt{5}-1\approx 1.236$. Because $1.236$ is not greater than $2$, we must reject this root.

We have searched every region and found no valid roots. Graphically, this means the function $f(t)=t(t-1)+2|t-1|+|t-2|$ never touches the $x$-axis; it sits entirely above it.

The set of real roots is the empty set. Never doubt your math—it is the most reliable compass you have.