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

Given,

$$x = 1 + \sqrt 2$$

$$=$$>  $$\frac{1}{x}=\frac{1}{1+\sqrt{2}}$$

$$=$$>  $$\frac{1}{x}=\frac{1}{1+\sqrt{2}}\times\frac{\sqrt{2}-1}{\sqrt{2}-1}$$

$$=$$>  $$\frac{1}{x}=\frac{\sqrt{2}-1}{2-1}$$

$$=$$>  $$\frac{1}{x}=\sqrt{2}-1$$

$$\therefore\ $$ $$\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2=x+\frac{1}{x}+2$$

$$=$$>  $$\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2=1+\sqrt{2}+\sqrt{2}-1+2$$

$$=$$>  $$\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2=2\sqrt{2}+2$$

$$=$$>  $$\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2=4.8284$$

$$=$$>  $$\sqrt{x}+\frac{1}{\sqrt{x}}=2.1973$$

Hence, the correct answer is Option B