1. Set up the equation

Let $a=\sqrt{3}+\sqrt{2}.$ Then $\sqrt{3}-\sqrt{2}=\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{1}{a},$ because $(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=3-2=1.$

So the given equation becomes $a^x+a^{-x}=10.$

2. Substitute a positive variable

Let $t=a^x.$ Since $a=\sqrt{3}+\sqrt{2}>1$, we have $t>0$ for all real $x$.

Then the equation becomes $t+\frac{1}{t}=10.$ Multiplying by $t$: $t^2-10t+1=0.$

3. Solve the quadratic

$t=\frac{10\pm\sqrt{100-4}}{2}=\frac{10\pm\sqrt{96}}{2}=\frac{10\pm 4\sqrt{6}}{2}=5\pm 2\sqrt{6}.$

Now observe: $5+2\sqrt{6}=(\sqrt{3}+\sqrt{2})^2=a^2,$ since $(\sqrt{3}+\sqrt{2})^2=3+2+2\sqrt{6}=5+2\sqrt{6}.$

Also, $5-2\sqrt{6}=(\sqrt{3}-\sqrt{2})^2=a^{-2}.$

So the two possible values of $t$ are $t=a^2 \quad \text{or} \quad t=a^{-2}.$

4. Find $x$

Since $t=a^x$,

• if $a^x=a^2$, then $x=2$;
• if $a^x=a^{-2}$, then $x=-2$.

Thus, $S=\{-2,2\}.$

So the number of elements in $S$ is $2.$

5. Check options

• A: $4$ ❌
• B: $0$ ❌
• C: $2$ ✅
• D: $1$ ❌

Hence the correct option is C.