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 the expression inside the square root: $7 + 4\sqrt{3} = 4 + 3 + 2(2)(\sqrt{3}) = (2 + \sqrt{3})^2

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$4$

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(A) Given $x = \sqrt{7 + 4\sqrt{3}}.$We can simplify the expression inside the square root: $7 + 4\sqrt{3} = 4 + 3 + 2(2)(\sqrt{3}) = (2 + \sqrt{3})^2.$Therefore,$x = \sqrt{(2 + \sqrt{3})^2} = 2 + \sqrt{3}.$Now,find $\frac{1}{x}:$$\frac{1}{x} = \frac{1}{2 + \sqrt{3}} = \frac{1}{2 + \sqrt{3}} 	imes \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}.$Finally,calculate $x + \frac{1}{x}:$$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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