Question

If $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3-\sqrt{2}}}$ find the value of $x^2+\frac{1}{x^2}$

• A. 38
• B. 98
• C. 62
• D. 47

Answer 1

Answer: (B)

98

$x+\frac{1}{x}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$=\frac{{(\sqrt{3}+\sqrt{2})}^2+{(\sqrt{3}-\sqrt{2})}^2}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}$
=$5\times 2=10$
$\therefore x^2+\frac{1}{x^2}={(x+\frac{1}{x})}^2-2$
$={10}^2-2=98$