Animated Solution for Mathematics - Basic Mathematics: The number of real roots of the equation $x|x-2|+3|x-3|+1=0$ is :

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an equation; we are performing a dissection. We are looking at the equation $x|x-2|+3|x-3|+1=0$.

At first glance, it looks like a standard algebraic puzzle, but the modulus bars are hiding a secret. They are not just symbols; they are switches. Depending on where you stand on the number line, they flip the sign of the expression inside.

Our goal is to map out these regions and see where the function truly hits zero.

Before we touch a single variable, we must define our territory. The modulus function $|f(x)|$ changes its behavior exactly when $f(x)=0$.

For our equation, we have two such points: $x-2=0$ (which gives $x=2$) and $x-3=0$ (which gives $x=3$). These are our 'critical points'.

They act as gates that divide the real number line into three distinct realms: 1. The region to the left of 2 ($x<2$) 2. The bridge between 2 and 3 ($2\le x<3$) 3. The vast expanse to the right of 3 ($x\ge 3$)

We must visit each realm separately to see if the equation holds true.

Imagine you are standing in the region where $x<2$. If you plug a value like $0$ into the expressions, both $(x-2)$ and $(x-3)$ yield negative results.

This means in this realm, the modulus bars must open with a negative sign to keep the overall value positive. Our equation transforms into:

Expanding this, we get $-x^2+2x-3x+9+1=0$, which simplifies to:

Using the quadratic formula, we find:

We have two candidates: $x\approx 2.70$ and $x\approx -3.70$. Since we are in the realm $x<2$, the value $2.70$ is an intruder and must be rejected. However, $x=\frac{-1-\sqrt{41}}{2}$ fits perfectly.

Now, we step into the bridge. Here, $(x-2)$ is positive, but $(x-3)$ is still negative. The equation becomes:

Expanding this, we get $x^2-2x-3x+9+1=0$, which simplifies to:

Let's check the discriminant:

A negative discriminant tells us that in this realm, the quadratic never touches the x-axis. There are no real roots here; the bridge is empty.

Finally, we reach the land of the positive. Here, both $(x-2)$ and $(x-3)$ are positive. The equation simplifies to:

Expanding, we get $x^2-2x+3x-9+1=0$, resulting in:

Using the quadratic formula, we get:

The candidates are $x\approx 2.37$ and $x\approx -3.37$. Since our condition is $x\ge 3$, neither of these values satisfies the constraint. They are both rejected.

We have traversed all three realms. We found one valid root in the first realm, zero in the second, and zero in the third.

The total count of real roots is exactly 1. This journey teaches us that in mathematics, the conditions we set (our intervals) are just as important as the calculations we perform. Never forget to validate your results against the constraints of your domain.