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 $34$

Rationalise $a=\frac{1}{3-2\sqrt{2}}$ as follows:

$a=\frac{1}{3-2\sqrt{2}}\times \frac{3+2\sqrt{2}}{3+2\sqrt{2}}=\frac{3+2\sqrt{2}}{{\left(3\right)}^2-{\left(2\sqrt{2}\right)}^2}=\frac{3+2\sqrt{2}}{9-8}=\frac{3+2\sqrt{2}}{1}=3+2\sqrt{2}$.

Rationalise $b=\frac{1}{3+2\sqrt{2}}$ as follows:

$b=\frac{1}{3+2\sqrt{2}}\times \frac{3-2\sqrt{2}}{3-2\sqrt{2}}=\frac{3-2\sqrt{2}}{{\left(3\right)}^2-{\left(2\sqrt{2}\right)}^2}=\frac{3-2\sqrt{2}}{9-8}=\frac{3-2\sqrt{2}}{1}=3-2\sqrt{2}$.

Now $a^2+b^2$ can be computed as:

$a^2+b^2={(3+2\sqrt{2})}^2+{(3-2\sqrt{2})}^2=9+8+12\sqrt{2}+9+8-12\sqrt{2}=9+8+9+8=34$.

Hence, $a^2+b^2=34$.

Therefore, option $A$ is correct.