Question

If $a=\frac{1}{3-2\sqrt{2}},b=\frac{1}{3+2\sqrt{2}}$, then the value of $a^2+b^2$ is __.

• A. $34$
• B. $35$
• C. $36$
• D. $37$

Answer 1

The correct option is A $34$

$a=\frac{1}{3-2\sqrt{2}}$
$a=\frac{3+2\sqrt{2}}{(3-2\sqrt{2})(3+2\sqrt{2})}$
$a=3+2\sqrt{2}$
$b=\frac{1}{3+2\sqrt{2}}$
$b=\frac{3-2\sqrt{2}}{(3-2\sqrt{2})(3+2\sqrt{2})}$
$b=3-2\sqrt{2}$
$a^2+b^2=(a+b{)}^2-2ab$
$a^2+b^2=(3+2\sqrt{2}+3-2\sqrt{2}{)}^2-2(3+2\sqrt{2}\left)\right(3-2\sqrt{2})$
$a^2+b^2=(6{)}^2-2$
$a^2+b^2=34$