Animated Solution for Mathematics - Quadratic Equations: Let $S=\{x\in R:(\sqrt{3}+\sqrt{2}{)}^x+(\sqrt{3}-\sqrt{2}{)}^x=10\}$. Then the number of elements in $S$ is :

Imagine you are standing before a mountain of a problem: $(\sqrt{3}+\sqrt{2}{)}^x+(\sqrt{3}-\sqrt{2}{)}^x=10$. At first glance, it looks like a chaotic mess of radicals and exponents.

In the world of JEE Advanced, whenever you see bases like $(\sqrt{3}+\sqrt{2})$ and $(\sqrt{3}-\sqrt{2})$, your algebraic spider-sense should tingle. These are conjugates.

When you multiply them, you get:

This is a beautiful, elegant property. It means that one base is simply the reciprocal of the other, which is the key that unlocks the entire problem.

Transforming the Exponential

Now, let us introduce a dummy variable to simplify our lives. Let $t=(\sqrt{3}+\sqrt{2}{)}^x$. Since the base is a positive number, any real power of it will also be positive, so $t>0$.

What happens to our second term? Since the base is the reciprocal of the first, raising it to the power $x$ is the same as $1/(\sqrt{3}+\sqrt{2}{)}^x$, which is simply $1/t$.

Our original, intimidating equation transforms into:

Solving for $t$

Now, multiply the entire equation by $t$ to clear the fraction. We get $t^2+1=10t$.

Rearranging this into standard form, we arrive at:

Let us use the quadratic formula to find $t$:

Connecting back to $x$

Here is the catch. We need to relate $t$ back to our original base. Look at $5\pm 2\sqrt{6}$.

Notice that $5=3+2$ and $6=3\times 2$. This perfectly matches the expansion of $(\sqrt{3}\pm \sqrt{2}{)}^2$. So, we can rewrite this expression beautifully as $(\sqrt{3}\pm \sqrt{2}{)}^2$.

For the positive case:

For the negative case:

The set of solutions is $S=\{2,-2\}$. The question asks for the number of elements in $S$, which is 2. You have conquered the mountain!