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

${(a+b)}^2=a^2+b^2+2ab{a}^2+b^2={(a+b)}^2-2aba=\frac{1}{3-2\sqrt{2}},b=\frac{1}{3+2\sqrt{2}}ab=\frac{1}{3-2\sqrt{2}}.\frac{1}{3+2\sqrt{2}}=\frac{1}{(3-2\sqrt{2})(3+2\sqrt{2})}=\frac{1}{9-8}=1a+b=\frac{1}{3-2\sqrt{2}}+\frac{1}{3+2\sqrt{2}}=\frac{3+2\sqrt{2}+3-2\sqrt{2}}{(3-2\sqrt{2})(3+2\sqrt{2})}=\frac{6}{1}=6{(a+b)}^2=6^2=36a^2+b^2={(a+b)}^2-2ab{a}^2+b^2=36-2=34.$