If x=7+4\sqrt{3}, then what is the value of \sqrt{x}+\frac{1}{\sqrt{x}}?

Explanation

Notice that x = 7+4\sqrt{3} = 4+3+2(2)(\sqrt{3}) = (2+\sqrt{3})^2. Thus, \sqrt{x} = 2+\sqrt{3}. Its reciprocal is \frac{1}{\sqrt{x}} = 2-\sqrt{3}. Adding them gives 2+\sqrt{3}+2-\sqrt{3} = 4.

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