If $$x=17-4\sqrt{18}$$, then find the value of $$(\sqrt{x}+\frac{1}{\sqrt{x}})$$  ?

Expression : $$x=17-4\sqrt{18}$$

=> $$x=17-2\sqrt{72}$$

=> $$x=(\sqrt9)^2+(\sqrt8)^2+2(\sqrt9)(\sqrt8)$$

Using, $$a^2+b^2+2ab=(a+b)^2$$

=> $$x=(\sqrt9+\sqrt8)^2$$

=> $$\sqrt{x}=3+2\sqrt2$$ -------------(i)

Also, $$\frac{1}{\sqrt{x}}=\frac{1}{3+2\sqrt2}$$

Rationalizing the denominator, we get :

=> $$\frac{1}{\sqrt{x}}=\frac{1}{3+2\sqrt2}\times(\frac{3-2\sqrt2}{3-2\sqrt2})$$

=> $$\frac{1}{\sqrt{x}}=\frac{3-2\sqrt2}{9-8}$$

=> $$\frac{1}{\sqrt{x}}=3-2\sqrt2$$ ---------(ii)

Adding equation (i) and (ii),

$$\therefore$$ $$(\sqrt{x}+\frac{1}{\sqrt{x}})=(3+2\sqrt2)+(3-2\sqrt2)=6$$

=> Ans - (D)