Question

If $x=3+\sqrt{8}$ then the value of $x^2+\frac{1}{x^2}$ is

• A. 30
• B. 32
• C. 34
• D. 36

Answer 1

Answer: (C)

34

$x=3+\sqrt{8}\Rightarrow \frac{1}{x}=\frac{1}{3+\sqrt{8}}\times \frac{3-\sqrt{8}}{3-\sqrt{8}}=\frac{3-\sqrt{8}}{9-8}$
$=3-\sqrt{8}$
$x^2+\frac{1}{x^2}={(x+\frac{1}{x})}^2-2=36-2=34$