Correct Answer: ans: 34.

Explanation:

Givenx = 3+$\sqrt{8}$……………(i)

=> $\frac{1}{x}$ = $\frac{1}{3+\sqrt{8}}$

=> $\frac{1}{x}$ =$\frac{3-\sqrt{8}}{(1+\sqrt{8})(3-\sqrt{8})}$

=> $\frac{1}{x}$ = $\frac{3-\sqrt{8}}{(3)^{2}-(\sqrt{8})^{2}}$

=> $\frac{1}{x}$ = $\frac{3-\sqrt{8}}{9-8}$

=> $\frac{1}{x}$ = $\frac{3-\sqrt{8}}{1}$

$\frac{1}{x}$ = (3-$\sqrt{8}$)……………………….(ii)

অতএব, (i) + (ii) হতে পাই

=> x+$\frac{1}{x}$ = (3+$(3+\sqrt{8)+(3-\sqrt{8})}$

=> x $\frac{1}{x}$ = 3+$\sqrt{8+3-\sqrt{8}}$

=> x $\frac{1}{x}$ = 6………………………………….(iii)

এখন, $x^{2}+\frac{1}{x^{2}}=(x+\frac{1}{x})^{2}-2x\times \frac{1}{x}$

= 6$^{2}$-2

= 36 -2

= 34