Find the positive square root
\[ \sqrt{7 + \sqrt{48}} \]
(a) \(7 + 2\sqrt{3}\) \quad (b) \(7 + \sqrt{3}\) \quad (c) \(2 + \sqrt{3}\) \quad (d) \(3 + \sqrt{2}\)
Solution:
\[ \sqrt{48} = 4\sqrt{3} \]
\[ \sqrt{7 + 4\sqrt{3}} = \sqrt{a} + \sqrt{b} \]
\[ a + b = 7,\quad ab = 12 \Rightarrow a = 3,\ b = 4 \]
\[ \therefore \sqrt{7 + \sqrt{48}} = \sqrt{3} + \sqrt{4} \]
\[ = \sqrt{3} + 2 = 2 + \sqrt{3} \]
\[ \therefore \text{Correct Answer: } 2 + \sqrt{3} \ (\text{Option c}) \]