Answer: B
\(\sqrt{x} + \sqrt{y} = \sqrt{18 + 6\sqrt{5}}\) Or, \(\sqrt{x} + \sqrt{y} = \sqrt{18 + 2\sqrt{45}}\) Or, \(\sqrt{x} + \sqrt{y} = \sqrt{(\sqrt{15})^2 + (\sqrt{3})^2 + 2 \cdot \sqrt{15} \cdot \sqrt{3}}\) Or, \(\sqrt{x} + \sqrt{y} = \sqrt{(\sqrt{15} + \sqrt{3})^2}\) Or, \(\sqrt{x} + \sqrt{y} = \sqrt{15} + \sqrt{3}\) \(\therefore x = 15\)
\(\sqrt{x} + \sqrt{y} = \sqrt{18 + 6\sqrt{5}}\) Or, \(\sqrt{x} + \sqrt{y} = \sqrt{18 + 2\sqrt{45}}\) Or, \(\sqrt{x} + \sqrt{y} = \sqrt{(\sqrt{15})^2 + (\sqrt{3})^2 + 2 \cdot \sqrt{15} \cdot \sqrt{3}}\) Or, \(\sqrt{x} + \sqrt{y} = \sqrt{(\sqrt{15} + \sqrt{3})^2}\) Or, \(\sqrt{x} + \sqrt{y} = \sqrt{15} + \sqrt{3}\) \(\therefore x = 15\)