Question

If $a=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and $b=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ the value of $a^2+b^2$ is

• A. 28
• B. 58
• C. 10
• D. 98

Answer 1

The correct option is D 98

$ab=1;$
$a+b=\frac{{(\sqrt{3}+\sqrt{2})}^2+{(\sqrt{3}-\sqrt{2})}^2}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}$
$=10$
$\therefore a^2+b^2={(a+b)}^2-2ab$
$={10}^2-2=98$