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

Navigating the Number Line

Welcome, future engineers. Today, we are not just solving an equation; we are embarking on a journey across the number line. We are looking at the equation $x|x+3|+|x-1|-2=0$.

At first glance, it looks like a standard algebraic problem, but the modulus signs—those vertical bars—are the true protagonists here. They are not just symbols; they are boundaries that represent a change in the very nature of the function. To solve this, we must become explorers of these boundaries.

The Boundary Patrol

Before we touch a single variable, we must identify where the 'identity crisis' of our modulus terms occurs. The modulus function $|u|$ is a shapeshifter: it is $u$ when $u\ge 0$, and $-u$ when $u<0$.

For the term $|x+3|$, the identity changes at $x=-3$. For the term $|x-1|$, it changes at $x=1$. These are our critical points.

They divide our universe into three distinct regions: the 'Left Zone' ($x<-3$), the 'Middle Zone' ($-3\le x<1$), and the 'Right Zone' ($x\ge 1$). We must visit each zone, solve the equation as it appears there, and then check if our solutions actually belong in that territory.

The Left Zone ($x<-3$)

Let us step into the first region: $x<-3$. In this cold, negative territory, both $(x+3)$ and $(x-1)$ are negative. Therefore, the modulus bars act as a negative multiplier.

We replace $|x+3|$ with $-(x+3)$ and $|x-1|$ with $-(x-1)$. Our equation transforms into:

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

Using the quadratic formula, we find:

Now, the verification: we assumed $x<-3$. The value $-2+\sqrt{3}\approx -0.27$ is clearly too large. However, $-2-\sqrt{3}\approx -3.73$ fits perfectly! We have found our first valid solution.

The Transition Zone ($-3\le x<1$)

Now, we move to the middle ground, where $x$ is greater than or equal to $-3$ but less than $1$. In this region, $(x+3)$ has turned positive, but $(x-1)$ is still negative. The equation becomes:

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

Applying the quadratic formula, we get:

Let's check our boundaries. $-1-\sqrt{2}\approx -2.41$ and $-1+\sqrt{2}\approx 0.41$. Both of these values fall comfortably within our interval of $[-3,1)$. We have just doubled our solution count!

The Right Zone ($x\ge 1$)

Finally, we reach the land of the positive: $x\ge 1$. Here, both expressions are positive. The modulus bars vanish, leaving us with:

This simplifies to $x^2+3x+x-1-2=0$, or:

Using the quadratic formula, we find:

We must be vigilant. We assumed $x\ge 1$. The value $-2+\sqrt{7}\approx 0.65$ is less than $1$, and $-2-\sqrt{7}$ is negative. Neither of these roots satisfies our condition, so this region yields no solutions.

We have traversed the entire number line. We found one solution in the Left Zone, two in the Middle Zone, and zero in the Right Zone. By methodically respecting the boundaries of the modulus, we have arrived at the truth: there are exactly 3 real solutions.

Remember, in JEE Advanced, the math is rarely just about the calculation; it is about the discipline of the process. You didn't just solve an equation; you mapped a function. Keep that clarity, keep that focus, and the solutions will always reveal themselves.