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).