Key Concepts and Formulas

• Conjugate Surds: Two surds of the form $a + \sqrt{b}$ and $a - \sqrt{b}$ are called conjugate surds. Their product is always a rational number: $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$.
• Reciprocal Relationship: If the product of two numbers is 1, they are reciprocals of each other. If $ab = 1$, then $a = \frac{1}{b}$ and $b = \frac{1}{a}$.
• Quadratic Formula: For a quadratic equation of the form $ax^2 + bx + c = 0$, the solutions are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

1. Identify the Reciprocal Relationship and Substitute

The given equation is: $(\sqrt 3 + \sqrt 2 )^{x^2 - 4} + (\sqrt 3 - \sqrt 2 )^{x^2 - 4} = 10$ We observe that $(\sqrt{3} + \sqrt{2})$ and $(\sqrt{3} - \sqrt{2})$ are conjugate surds. Let's find their product: $(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$ Since their product is 1, they are reciprocals of each other. This means: $\sqrt{3} - \sqrt{2} = \frac{1}{\sqrt{3} + \sqrt{2}} = (\sqrt{3} + \sqrt{2})^{-1}$ Now, let $y = (\sqrt{3} + \sqrt{2})^{x^2 - 4}$. This substitution will simplify the equation. Using the reciprocal relationship, we can rewrite the second term: $(\sqrt{3} - \sqrt{2})^{x^2 - 4} = \left( (\sqrt{3} + \sqrt{2})^{-1} \right)^{x^2 - 4} = (\sqrt{3} + \sqrt{2})^{-(x^2 - 4)} = \frac{1}{(\sqrt{3} + \sqrt{2})^{x^2 - 4}} = \frac{1}{y}$ Substituting these into the original equation, we get: $y + \frac{1}{y} = 10$

2. Solve the Quadratic Equation for $y$

To solve for $y$, we multiply the entire equation by $y$ (since $y$ is an exponential term and thus positive, $y \neq 0$): $y^2 + 1 = 10y$ Rearrange the equation into standard quadratic form: $y^2 - 10y + 1 = 0$ Now, we use the quadratic formula to find the values of $y$. Here, $a=1$, $b=-10$, and $c=1$: $y = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(1)}}{2(1)}$ $y = \frac{10 \pm \sqrt{100 - 4}}{2}$ $y = \frac{10 \pm \sqrt{96}}{2}$ Simplify the radical $\sqrt{96}$: $\sqrt{96} = \sqrt{16 \times 6} = \sqrt{16} \times \sqrt{6} = 4\sqrt{6}$ Substitute this back into the expression for $y$: $y = \frac{10 \pm 4\sqrt{6}}{2}$ Divide by 2: $y = 5 \pm 2\sqrt{6}$ We now have two possible values for $y$: $y_1 = 5 + 2\sqrt{6}$ and $y_2 = 5 - 2\sqrt{6}$.

3. Back-Substitute and Solve for $x$ (Case 1)

We substitute $y$ back into our original substitution $y = (\sqrt{3} + \sqrt{2})^{x^2 - 4}$.

Case 1: $y = 5 + 2\sqrt{6}$ We need to express $5 + 2\sqrt{6}$ with a base of $(\sqrt{3} + \sqrt{2})$. Recall that $(\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab}$. Let's check if $5 + 2\sqrt{6}$ matches this form: $(\sqrt{3} + \sqrt{2})^2 = (\sqrt{3})^2 + (\sqrt{2})^2 + 2(\sqrt{3})(\sqrt{2}) = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6}$ Thus, $5 + 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^2$. Now, substitute this back into our expression for $y$: $(\sqrt{3} + \sqrt{2})^{x^2 - 4} = (\sqrt{3} + \sqrt{2})^2$ Since the bases are equal, the exponents must be equal: $x^2 - 4 = 2$ $x^2 = 6$ Taking the square root of both sides: $x = \pm \sqrt{6}$ These are two real solutions for $x$.

4. Back-Substitute and Solve for $x$ (Case 2)

Case 2: $y = 5 - 2\sqrt{6}$ Similarly, we need to express $5 - 2\sqrt{6}$ with a base of $(\sqrt{3} + \sqrt{2})$. We know that $5 - 2\sqrt{6} = (\sqrt{3} - \sqrt{2})^2$. Since our base is $(\sqrt{3} + \sqrt{2})$, we use the reciprocal relationship: $(\sqrt{3} - \sqrt{2}) = (\sqrt{3} + \sqrt{2})^{-1}$. Therefore: $5 - 2\sqrt{6} = ((\sqrt{3} + \sqrt{2})^{-1})^2 = (\sqrt{3} + \sqrt{2})^{-2}$ Now, substitute this back into our expression for $y$: $(\sqrt{3} + \sqrt{2})^{x^2 - 4} = (\sqrt{3} + \sqrt{2})^{-2}$ Since the bases are equal, the exponents must be equal: $x^2 - 4 = -2$ $x^2 = 2$ Taking the square root of both sides: $x = \pm \sqrt{2}$ These are another two real solutions for $x$.

5. Determine the Number of Solutions $n(S)$

The set of all possible values for $x$ is $S = \{ \sqrt{6}, -\sqrt{6}, \sqrt{2}, -\sqrt{2} \}$. All these values are real and distinct. Therefore, the number of elements in set $S$, denoted as $n(S)$, is 4.

Common Mistakes & Tips

• Forgetting the Reciprocal: When dealing with conjugate surds, always remember the reciprocal relationship. This is key to simplifying the equation.
• Missing Negative Roots: When solving $x^2 = k$, remember that there are two solutions: $x = \pm \sqrt{k}$.
• Base Matching: Before equating exponents, make sure the bases are identical. Use the reciprocal relationship to adjust if necessary.

Summary

To solve the given exponential equation, we recognized the reciprocal relationship between the bases, used a substitution to transform the equation into a quadratic, solved for the substituted variable, and then back-substituted to find the values of $x$. We found four distinct real values for $x$: $\sqrt{6}, -\sqrt{6}, \sqrt{2},$ and $-\sqrt{2}$. Thus, $n(S) = 4$.

Final Answer

The final answer is $\boxed{4}$, which corresponds to option (B).