Let S = { x : x in R text{ and } (sqrt{3} + s | vedclass

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(B) Let $t = (\sqrt{3} + \sqrt{2})^{x^2 - 4}$.Since $(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$,we have $(\sqrt{3} - \sqrt{2}) = \frac{1}

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$4$

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(B) Let $t = (\sqrt{3} + \sqrt{2})^{x^2 - 4}$.Since $(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$,we have $(\sqrt{3} - \sqrt{2}) = \frac{1}{\sqrt{3} + \sqrt{2}}$.Thus,the equation becomes $t + \frac{1}{t} = 10$.$t^2 - 10t + 1 = 0$.Using the quadratic formula,$t = \frac{10 \pm \sqrt{100 - 4}}{2} = 5 \pm \sqrt{24} = 5 \pm 2\sqrt{6}$.Note that $5 + 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^2$ and $5 - 2\sqrt{6} = (\sqrt{3} - \sqrt{2})^2 = (\sqrt{3} + \sqrt{2})^{-2}$.Case $1$: $(\sqrt{3} + \sqrt{2})^{x^2 - 4} = (\sqrt{3} + \sqrt{2})^2 \implies x^2 - 4 = 2 \implies x^2 = 6 \implies x = \pm \sqrt{6}$.Case $2$: $(\sqrt{3} + \sqrt{2})^{x^2 - 4} = (\sqrt{3} + \sqrt{2})^{-2} \implies x^2 - 4 = -2 \implies x^2 = 2 \implies x = \pm \sqrt{2}$.Thus,$S = \{ \sqrt{6}, -\sqrt{6}, \sqrt{2}, -\sqrt{2} \}$.The number of elements $n(S) = 4$.

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

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