If n = 7 + 4 sqrt{3},then the value of left(s | vedclass

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(B) Given $n = 7 + 4 \sqrt{3}$.We can rewrite $n$ as $n = 4 + 3 + 2 	imes 2 	imes \sqrt{3} = 2^2 + (\sqrt{3})^2 + 2(2)(\sqrt{3})$.Using the identity

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$4$

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(B) Given $n = 7 + 4 \sqrt{3}$.We can rewrite $n$ as $n = 4 + 3 + 2 	imes 2 	imes \sqrt{3} = 2^2 + (\sqrt{3})^2 + 2(2)(\sqrt{3})$.Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$,we get $n = (2 + \sqrt{3})^2$.Therefore,$\sqrt{n} = 2 + \sqrt{3}$.Now,find $\frac{1}{\sqrt{n}} = \frac{1}{2 + \sqrt{3}}$.Rationalizing the denominator: $\frac{1}{2 + \sqrt{3}} 	imes \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$.Finally,$\sqrt{n} + \frac{1}{\sqrt{n}} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$.

If $n = 7 + 4 \sqrt{3}$,then the value of $\left(\sqrt{n} + \frac{1}{\sqrt{n}}\right)$ is

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