Animated Solution for Mathematics - Quadratic Equations: Let $S=\{x:x\in \mathbb{R}\text{ and }(\sqrt{3}+\sqrt{2}{)}^{x^2-4}+(\sqrt{3}-\sqrt{2}{)}^{x^2-4}=10\}$. Then $n(S)$ is equal to

Imagine you are standing before a complex exponential equation:

At first glance, it looks like a nightmare of radicals and quadratic exponents. But in the world of JEE Advanced, complexity is often just a mask for hidden simplicity.

The first step in our journey is to observe the bases: $(\sqrt{3}+\sqrt{2})$ and $(\sqrt{3}-\sqrt{2})$. These are not random numbers; they are conjugates.

When we multiply them, we get:

This is the master key. It tells us that:

From Exponential to Algebraic

Now that we have this insight, let's simplify the problem. We introduce a dummy variable, $t$, defined as:

Because of the reciprocal relationship we just discovered, the second term in our equation, $(\sqrt{3}-\sqrt{2}{)}^{x^2-4}$, becomes exactly $\frac{1}{t}$.

Our equation now transforms into the much friendlier form:

This is the magic of substitution. We have stripped away the intimidating exponential layer to reveal a simple algebraic structure.

To solve $t+\frac{1}{t}=10$, we multiply by $t$ to get $t^2+1=10t$, which rearranges into the quadratic equation:

Using the quadratic formula, $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we find:

Since $\sqrt{96}=4\sqrt{6}$, we simplify to:

We are now in the home stretch. We need to relate these values of $t$ back to our original base, $(\sqrt{3}+\sqrt{2})$.

Notice that:

Similarly, $(\sqrt{3}-\sqrt{2}{)}^2=5-2\sqrt{6}$, which is equivalent to $(\sqrt{3}+\sqrt{2}{)}^{-2}$.

Case 1: Where $t=5+2\sqrt{6}$, we have:

Equating the exponents, $x^2-4=2$, so $x^2=6$, giving $x=\pm \sqrt{6}$.

Case 2: Where $t=5-2\sqrt{6}$, we have:

Equating the exponents, $x^2-4=-2$, so $x^2=2$, giving $x=\pm \sqrt{2}$.

We have found four distinct real solutions: $\{-\sqrt{6},-\sqrt{2},\sqrt{2},\sqrt{6}\}$. The number of solutions is 4.