Root these pairs

Square both sides of the equation:

$a+b+2\sqrt{ab}=2+\sqrt{3}$

If $a$ and $b$ are rational numbers, then $a+b=2$ and $2\sqrt{ab}=\sqrt{3}$ .

Solving for $b$ in the second equation:

$\sqrt{ab}=\frac{\sqrt{3}}{2}$

$ab=\frac{3}{4}$

$b=\frac{3}{4a}$

Substitute this into the first equation:

$a+\frac{3}{4a}=2$

$4a^2+3=8a$

$4a^2-8a+3=0$

$(2a-1)(2a-3)=0$

Thus, $a=\frac{1}{2}$ or $a=\frac{3}{2}$ . These correspond to $b=\frac{3}{2}$ and $b=\frac{1}{2}$ , respectively.

Multiply both sides of the equation by $\sqrt{2}$

$\sqrt{2a}+\sqrt{2b} = \sqrt{4+2\sqrt{3}}$

$\sqrt{2a}+\sqrt{2b} = \sqrt{3} + 1$

Therefore possible values of (a,b) are ( $\frac{1}{2},\frac{3}{2}$ ). and ( $\frac{3}{2},\frac{1}{2}$ ) that is $\boxed{2}$ possible values.