Animated Solution for Mathematics - Quadratic Equations: The number of real roots of the equation $\sqrt{x^2-4x+3}+\sqrt{x^2-9}=\sqrt{4x^2-14x+6}$, is:

My dear student, welcome to the battlefield of JEE Advanced. When you see an equation like:

Your first instinct might be to panic or, worse, to start squaring everything in sight. Please, take a deep breath. In the world of competitive mathematics, brute force is rarely the path to victory. Elegance is.

The first step in any radical equation is to look for the 'DNA' of the problem. We have three quadratic expressions under square roots. If you try to square this immediately, you will create a monster of an equation that will haunt your scratchpad.

Instead, let us factorize:

The first term: $x^2-4x+3=(x-1)(x-3)$ The second term: $x^2-9=(x-3)(x+3)$ * The right-hand side: $4x^2-14x+6=2(2x^2-7x+3)=2(2x-1)(x-3)$

Do you see it? The factor $(x-3)$ is the heartbeat of this equation. It appears in every single term. This is not a coincidence; it is a gift from the problem setter.

Your Green Region

Before we solve, we must define where our solution is allowed to exist. In the realm of real numbers, the expression under a square root must be non-negative. This is our 'domain constraint.'

For the first term, $(x-1)(x-3)\ge 0$, which means $x\in (-\infty ,1]\cup [3,\infty )$.

For the second term, $(x-3)(x+3)\ge 0$, which means $x\in (-\infty ,-3]\cup [3,\infty )$.

For the third term, $(2x-1)(x-3)\ge 0$, which means $x\in (-\infty ,\frac{1}{2}]\cup [3,\infty )$.

To find the valid playground, we need the intersection of these sets. If you draw this on a number line, you will see that the only regions where all three conditions are satisfied simultaneously are $x\in (-\infty ,-3]\cup [3,\infty )$. This is our green region. Any answer we find outside this zone is an impostor.

Now, let us test the boundary point $x=3$. If we plug $x=3$ into our factored equation, every term becomes zero. $0+0=0$. It works! We have found our first root.

Now, what about the rest of the domain? We split this into two cases.

Case 1: $x>3$ Here, $(x-3)$ is strictly positive. We can safely divide the entire equation by $\sqrt{x-3}$. This leaves us with:

Now, squaring is much safer. After squaring both sides, we get:

Simplifying this leads us to $2\sqrt{x^2+2x-3}=2x-4$, or $\sqrt{x^2+2x-3}=x-2$. Squaring again, we get $x^2+2x-3=x^2-4x+4$, which simplifies to $6x=7$, or $x=\frac{7}{6}$.

But wait! We assumed $x>3$. Since $\frac{7}{6}$ is approximately $1.16$, it is not greater than $3$. We must reject it.

Case 2: $x\le -3$ Here, $(x-3)$ is negative. We handle this by factoring out a negative sign and dividing by $\sqrt{3-x}$. Following the same algebraic dance, we again arrive at $x=\frac{7}{6}$. Again, this fails our condition $x\le -3$.

We have exhausted all possibilities. The only survivor of our rigorous domain check is $x=3$.

The JEE Advanced exam loves to test whether you can distinguish between algebraic solutions and valid domain solutions. Never forget: the algebra is the tool, but the domain is the law. Keep practicing, keep questioning, and keep falling in love with the logic.