Let S=\xin R:(sqrt3+sqrt2)^x+(sqrt3-sqrt2)^x=10\$S=\{x\in R:(\sqrt{3}+\sqrt{2})^{x}+(\sqrt{3}-\sqrt{2})^{x}=10\}$. Then the number of elements in S is:

### Related Formula Conjugate Surd Identity: (sqrta + sqrtb)(sqrta - sqrtb) = a - b$$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$$ ### Core Logic Observe the base components of the exponents: (sqrt3 + sqrt2)(sqrt3 - sqrt2) = 3 - 2 = 1$$(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$$ Therefore, we can express one base as the reciprocal of the other: sqrt3 - sqrt2 = frac1sqrt3 + sqrt2$$\sqrt{3} - \sqrt{2} = \frac{1}{\sqrt{3} + \sqrt{2}}$$ The equation becomes: (sqrt3 + sqrt2)^x + frac1(sqrt3 + sqrt2)^x = 10 $$(\sqrt{3} + \sqrt{2})^{x} + \frac{1}{(\sqrt{3} + \sqrt{2})^{x}} = 10 $$ ### Step 1: Formulate the Quadratic Equation Let (sqrt3 + sqrt2)^x = t$(\sqrt{3} + \sqrt{2})^{x} = t$. Then: t + frac1t = 10 implies t^2 - 10t + 1 = 0 $$t + \frac{1}{t} = 10 \implies t^{2} - 10t + 1 = 0 $$ Solving for t$t$ using the quadratic formula: t = frac10 pm sqrt(-10)^2 - 4(1)(1)2 = frac10 pm sqrt962 = 5 pm 2sqrt6 $$t = \frac{10 \pm \sqrt{(-10)^{2} - 4(1)(1)}}{2} = \frac{10 \pm \sqrt{96}}{2} = 5 \pm 2\sqrt{6} $$ ### Step 2: Solve for x Notice that (5 pm 2sqrt6)$(5 \pm 2\sqrt{6})$ can be written as square powers of the original base: (sqrt3 pm sqrt2)^2 = 3 + 2 pm 2sqrt6 = 5 pm 2sqrt6$$(\sqrt{3} \pm \sqrt{2})^{2} = 3 + 2 \pm 2\sqrt{6} = 5 \pm 2\sqrt{6}$$ Thus, we have: - For t = 5 + 2sqrt6 implies (sqrt3 + sqrt2)^x = (sqrt3 + sqrt2)^2 implies x = 2$t = 5 + 2\sqrt{6} \implies (\sqrt{3} + \sqrt{2})^{x} = (\sqrt{3} + \sqrt{2})^{2} \implies x = 2$ - For t = 5 - 2sqrt6 implies (sqrt3 + sqrt2)^x = (sqrt3 - sqrt2)^2 = (sqrt3 + sqrt2)^-2 implies x = -2$t = 5 - 2\sqrt{6} \implies (\sqrt{3} + \sqrt{2})^{x} = (\sqrt{3} - \sqrt{2})^{2} = (\sqrt{3} + \sqrt{2})^{-2} \implies x = -2$ Therefore, the distinct real solutions are x = 2$x = 2$ and x = -2$x = -2$. The number of elements in set S$S$ is 2$2$. ### Pattern Recognition Sees: Rational conjugate bases added with inverse matching variables. Shortcut: Whenever you see an equation of the form A^x + B^x = C$A^x + B^x = C$ where AB = 1$AB = 1$, the solution will always be symmetric (pm x_0$\pm x_0$). Checking x=2$x=2$ explicitly gives (sqrt3+sqrt2)^2 + (sqrt3-sqrt2)^2 = (5+2sqrt6) + (5-2sqrt6) = 10$(\sqrt{3}+\sqrt{2})^2 + (\sqrt{3}-\sqrt{2})^2 = (5+2\sqrt{6}) + (5-2\sqrt{6}) = 10$, confirming pm 2$\pm 2$ immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Quadratic Equations Class 9 Mathematics: Number Systems (Rationalization)