If x = 3 + 8 , then x^2 + 1x^2 is equal to = ?

Answer: Option B

$$\begin{array}{rl}& \because x=3+\sqrt{8}\\ & \Rightarrow x^2={\left(3+\sqrt{8}\right)}^2\\ & \Rightarrow x^2=3^2+{\left(\sqrt{8}\right)}^2+2\times 3\times \sqrt{8}\\ & \Rightarrow x^2=9+8+6\sqrt{8}\\ & \Rightarrow x^2=17+12\sqrt{2}\end{array}$$
$$\therefore x^2+\frac{1}{x^2}$$
$$=\left(17+12\sqrt{2}\right)$$   $$+\frac{1}{\left(17+12\sqrt{2}\right)}$$   $$\times \frac{\left(17-12\sqrt{2}\right)}{\left(17-12\sqrt{2}\right)}$$
$$\begin{array}{rl}& =\left(17+12\sqrt{2}\right)+\frac{\left(17-12\sqrt{2}\right)}{289-288}\\ & =\left(17+12\sqrt{2}\right)+\left(17-12\sqrt{2}\right)\\ & =17+17\\ & =34\end{array}$$