If x = sqrt{7 + 4sqrt{3}},then x + frac{1}{x} | vedclass

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(A) Given $x = \sqrt{7 + 4\sqrt{3}}$.We can simplify $x$ as follows:$x = \sqrt{4 + 3 + 2(2)(\sqrt{3})} = \sqrt{2^2 + (\sqrt{3})^2 + 2(2)(\sqrt{3})} =

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$4$

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(A) Given $x = \sqrt{7 + 4\sqrt{3}}$.We can simplify $x$ as follows:$x = \sqrt{4 + 3 + 2(2)(\sqrt{3})} = \sqrt{2^2 + (\sqrt{3})^2 + 2(2)(\sqrt{3})} = \sqrt{(2 + \sqrt{3})^2} = 2 + \sqrt{3}$.Now,find $\frac{1}{x}$:$\frac{1}{x} = \frac{1}{2 + \sqrt{3}} = \frac{1}{2 + \sqrt{3}} 	imes \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$.Therefore,$x + \frac{1}{x} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$.

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

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