Notice that if both [tex]x,y[/tex] are positive, then
[tex]xy = 3 \implies \sqrt{xy} = \sqrt x \sqrt y = \sqrt3[/tex]
We also have the binomial expansion
[tex]x + 2\sqrt{x}\sqrt{y} + y = \left(\sqrt x\right)^2 + 2 \sqrt x \sqrt y + \left(\sqrt y\right)^2 = \left(\sqrt x + \sqrt y\right)^2[/tex]
Then
[tex]\left(\sqrt x + \sqrt y\right)^2 = (x + y) + 2\sqrt{xy} = 5 + 2\sqrt3 \\\\ \implies \sqrt x + \sqrt y = \boxed{\sqrt{5 + 2\sqrt3}}[/tex]
Let's see if we can denest the radical. Suppose we could write
[tex]\sqrt{5 + 2\sqrt3} = a + b \sqrt c[/tex]
for some non-zero integer constants [tex]a,b,c[/tex] (and [tex]c>0[/tex]). Squaring both sides gives
[tex]5 + 2\sqrt3 = a^2 + 2ab\sqrt c + b^2c[/tex]
Let [tex]c=3[/tex] and [tex]ab=1[/tex]. Then [tex]a^2 + b^2c = 5[/tex], which gives
[tex]a^2 + 3b^2 = 5[/tex]
[tex]a^2 + \dfrac3{a^2} = 5[/tex]
[tex]a^4 - 5a^2 + 3 = 0[/tex]
We can solve this with the quadratic formula. However, it'll lead to non-integer solutions for [tex]a,b[/tex], so we cannot denest after all.
