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

$a^2+b^2=(a+b{)}^2-2ab$

$a^2+b^2=(\frac{1}{3-2\sqrt{2}}+\frac{1}{3+2\sqrt{2}}{)}^2-2\times \frac{1}{3-2\sqrt{2}}\times \frac{1}{3+2\sqrt{2}}$

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

$a^2+b^2=6^2-2$

$a^2+b^2=34$