Speed test. Solve as fast as you can

For this to simplify, we know that $3+2\sqrt{2}$ can be rewritten in the form $(x+\sqrt{y})^{2}$ .

So $3+2\sqrt{2}=(x+\sqrt{y})^{2}$ for some rational numbers $x, y$ .

Expanding the RHS, we have: $3+2\sqrt{2}=x^{2}+2x\sqrt{y}+y$ .

Since $2x\sqrt{y}$ is the only irrational term on the RHS and \(2\sqrt{2})\ is the only irrational term on the LHS, we know they have to be equal.

We have: \(2x\sqrt{y}=2\sqrt{2}\) which leads to $x=1$ and $y=2$ .

So $3+2\sqrt{2}=(1+\sqrt{2})^{2}$ .

Let's plug this back in to the problem.

$\sqrt{3+2\sqrt{2}}$

$= \sqrt{(1+\sqrt{2})^{2}}$

$=1+\sqrt{2}$ .

So $a=1, b=2$ . This means that our answer is $\frac{b}{\sqrt{a}}=\frac{2}{\sqrt{1}}=\frac{2}{1}=\boxed{2}$ .

This took me a grand total of 5 seconds. (Mathcounts countdown round practice haha)

can simply to 1 + 2^(1/2) --> 2/1^(1/2) = 2/1 = 2