Question : If $n= 7+4\sqrt{3}$, the value of $(\sqrt{n}+\frac{1}{\sqrt{n}})$ is:
Option 1: $2\sqrt{3}$
Option 2: $4$
Option 3: $-4$
Option 4: $-2\sqrt{3}$
Correct Answer: $4$
Solution :
Given: $n= 7+4\sqrt{3}$
So, $\sqrt{n}=\sqrt{7+4\sqrt{3}}=\sqrt{2^2+(\sqrt{3})^2+2×2×\sqrt{3}}=\sqrt{(2+\sqrt{3})^2}=2+\sqrt{3}$
Therefore, $\frac{1}{\sqrt{n}}=\frac{1}{(2\:+\:\sqrt{3})}×\frac{(2\:-\:\sqrt{3})}{(2\:-\:\sqrt{3})}=(2-\sqrt{3})$
So, $(\sqrt{n}+\frac{1}{\sqrt{n}})=2+\sqrt{3}+2-\sqrt{3}=4$
Hence, the correct answer is $4$.
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