Let S=x R:(√3+√2)^x+(√3-√2)^x=10.

Question

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

Accepted Answer

Notice that $(\sqrt{3}+\sqrt{2})$ and $(\sqrt{3}-\sqrt{2})$ are reciprocals of each other because :

$(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=3-2=1$

Using the Reciprocal Property :

This means $(\sqrt{3}-\sqrt{2}{)}^{x}=\frac{1}{(\sqrt{3}+\sqrt{2}{)}^{x}}$

Let $a=(\sqrt{3}+\sqrt{2}{)}^{x}$. The equation given in the problem becomes :

$a+\frac{1}{a}=10$

Simplifying

Multiplying both sides by *a*, we get a quadratic equation :

${a}^{2}+1=10a$

${a}^{2}-10a+1=0$

Solving the Quadratic

Using the quadratic formula, we find :

$a=\frac{10\pm \sqrt{96}}{2}=5\pm 2\sqrt{6}$

Possible values of x

Since $a=(\sqrt{3}+\sqrt{2}{)}^{x}$, we have two cases :

$(\sqrt{3}+\sqrt{2}{)}^{x}=5+2\sqrt{6}$ = $(\sqrt{3}+\sqrt{2}{)}^{2}$. There is one real solution for x in this case which is x = 2.

$(\sqrt{3}+\sqrt{2}{)}^{x}=5-2\sqrt{6}$ = $(\sqrt{3}-\sqrt{2}{)}^{2}=\frac{1}{(\sqrt{3}+\sqrt{2}{)}^{2}}$ = $(\sqrt{3}+\sqrt{2}{)}^{-2}$. There is one real solution for x in this case which is x = - 2.

Conclusion

There are two real solutions for *x*. Therefore, the number of elements in the set S is 2. This corresponds with option (C).

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