If $(a +b)^2 = 5 + 2\sqrt{6}$, what can be the possible value of ‘b’ from the following?

$\sqrt{7}$ $\sqrt{3}$ $\sqrt{6}$ 5

$\sqrt{3}$

$(a +b)^2 = 5 + 2\sqrt{6}$ ⇒ $(a +b)^2 = (2) + (3) + 2\sqrt{2×3}$ ⇒ (a)2 + (b)2 + 2(a×b) = (\(\sqrt {2}\))2 + (\(\sqrt {3}\))2 + 2\(\sqrt {2×3}\) ⇒ a or b can be = $\sqrt{2}$ or $\sqrt{3}$ ⇒ $(\sqrt{2}+\sqrt{3})^2=5 + 2\sqrt{6}$ Hence, $b=\sqrt{3}$