If $ x = 1 + \sqrt{2}$, then find the value of $\sqrt{x} +(\frac{1}{\sqrt{x}})$.

2.1014 2.1973 1.9876 1.9996

2.1973

If $ x = 1 + \sqrt{2}$, then find the value of $\sqrt{x} +(\frac{1}{\sqrt{x}})$ If $ x = 1 + \sqrt{2}$ then, x = 2.41  ( because $\sqrt{2}$ = 1.41 ) $\sqrt{x} +(\frac{1}{\sqrt{x}})$ = \(\frac{x + 1}{\sqrt{x}}\) = \(\frac{2.41 + 1}{2.41}\) = \(\frac{3.41 }{1.55}\) = 2.1973