If x = (7 - 4sqrt{3}),then the value of left( | vedclass

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(C) Given $x = 7 - 4\sqrt{3}$.First,find the value of $\frac{1}{x}$ by rationalizing the denominator:$\frac{1}{x} = \frac{1}{7 - 4\sqrt{3}} = \frac{1

Vedclass · Accepted Answer

$14$

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(C) Given $x = 7 - 4\sqrt{3}$.First,find the value of $\frac{1}{x}$ by rationalizing the denominator:$\frac{1}{x} = \frac{1}{7 - 4\sqrt{3}} = \frac{1 	imes (7 + 4\sqrt{3})}{(7 - 4\sqrt{3}) 	imes (7 + 4\sqrt{3})}$Using the identity $(a - b)(a + b) = a^2 - b^2$:$\frac{1}{x} = \frac{7 + 4\sqrt{3}}{7^2 - (4\sqrt{3})^2} = \frac{7 + 4\sqrt{3}}{49 - (16 	imes 3)} = \frac{7 + 4\sqrt{3}}{49 - 48} = \frac{7 + 4\sqrt{3}}{1} = 7 + 4\sqrt{3}$Now,calculate $x + \frac{1}{x}$:$x + \frac{1}{x} = (7 - 4\sqrt{3}) + (7 + 4\sqrt{3})$$x + \frac{1}{x} = 7 + 7 - 4\sqrt{3} + 4\sqrt{3} = 14$.

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

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