Solve this
Question:

If $x=3+\sqrt{8}$ then $\left(x^{2}+\frac{1}{x^{2}}\right)=?$

(a) 34

(b) 56

(c) 28

(d) 63

Solution:

Given: $x=3+\sqrt{8}$

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

$x+\frac{1}{x}=(3+\sqrt{8})+(3-\sqrt{8})=6$

$\left(x+\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}+2 \times x \times \frac{1}{x}=x^{2}+\frac{1}{x^{2}}+2$

$\Rightarrow 6^{2}=x^{2}+\frac{1}{x^{2}}+2$

$\Rightarrow 36=x^{2}+\frac{1}{x^{2}}+2$

$\Rightarrow x^{2}+\frac{1}{x^{2}}=36-2=34$

Hence, the correct answer is option (a).