1. Set up the expression

We need to solve

$$(\sqrt3+\sqrt2)^{x^2-4}+(\sqrt3-\sqrt2)^{x^2-4}=10.$$

Let

$$a=\sqrt3+\sqrt2.$$

Then

$$\sqrt3-\sqrt2=\frac{1}{\sqrt3+\sqrt2}=\frac{1}{a},$$

because

$$(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)=3-2=1.$$

So the equation becomes

$$a^{x^2-4}+a^{-(x^2-4)}=10.$$

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2. Substitute the exponent

Let

$$t=x^2-4.$$

Then we need to solve

$$a^t+a^{-t}=10.$$

Now compute a useful value:

$$a^2=(\sqrt3+\sqrt2)^2=3+2+2\sqrt6=5+2\sqrt6.$$

Also,

$$a^{-2}=(\sqrt3-\sqrt2)^2=3+2-2\sqrt6=5-2\sqrt6.$$

Hence

$$a^2+a^{-2}=(5+2\sqrt6)+(5-2\sqrt6)=10.$$

So $t=2$ is a solution.

Also, since the expression is symmetric in $t$ and $-t$,

$$a^{-2}+a^2=10,$$

so $t=-2$ is also a solution.

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3. Show these are the only solutions for $t$

Let

$$y=a^t>0.$$

Then the equation becomes

$$y+\frac1y=10.$$

Multiplying by $y$,

$$y^2-10y+1=0.$$

Solving,

$$y=\frac{10\pm\sqrt{100-4}}{2}=\frac{10\pm\sqrt{96}}{2}=5\pm2\sqrt6.$$

But

$$5+2\sqrt6=a^2, \qquad 5-2\sqrt6=a^{-2}.$$

Thus

$$a^t=a^2 \quad \text{or} \quad a^t=a^{-2}.$$

Since $a=\sqrt3+\sqrt2>1$, the exponential function is one-one, so

$$t=2 \quad \text{or} \quad t=-2.$$

Thus,

$$x^2-4=2 \quad \text{or} \quad x^2-4=-2.$$

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4. Solve for $x$

From

$$x^2-4=2,$$

we get

$$x^2=6 \implies x=\pm\sqrt6.$$

From

$$x^2-4=-2,$$

we get

$$x^2=2 \implies x=\pm\sqrt2.$$

So

$$S=\{\pm\sqrt6,\,\pm\sqrt2\}.$$

Therefore,

$$n(S)=4.$$

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5. Check options

• A: $6$
• B: $4$
• C: $0$
• D: $2$

Hence the correct option is B.