If $x=7-4\sqrt{3}$, then $\sqrt{x}+\frac{1}{\sqrt{x}}=\text{?}$

Given $x=7-4\sqrt{3}$.
We can express $x$ as a perfect square:

$$x=4+3-2(2)(\sqrt{3})=2^2+(\sqrt{3}{)}^2-2(2)(\sqrt{3})=(2-\sqrt{3}{)}^2$$

Taking the square root on both sides:

$$\sqrt{x}=2-\sqrt{3}$$

Now, find $\frac{1}{\sqrt{x}}$ by rationalizing the denominator:

$$\frac{1}{\sqrt{x}}=\frac{1}{2-\sqrt{3}}=\frac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}$$

Now, compute $\sqrt{x}+\frac{1}{\sqrt{x}}$:

$$\sqrt{x}+\frac{1}{\sqrt{x}}=(2-\sqrt{3})+(2+\sqrt{3})=4$$