If x = 7 - 4sqrt{3},then find the value of x^ | vedclass

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(D) Given $x = 7 - 4\sqrt{3}$.First,find $\frac{1}{x}$:$\frac{1}{x} = \frac{1}{7 - 4\sqrt{3}}$.Rationalizing the denominator:$\frac{1}{x} = \frac{1}{7

Vedclass · Accepted Answer

$194$

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(D) Given $x = 7 - 4\sqrt{3}$.First,find $\frac{1}{x}$:$\frac{1}{x} = \frac{1}{7 - 4\sqrt{3}}$.Rationalizing the denominator:$\frac{1}{x} = \frac{1}{7 - 4\sqrt{3}} 	imes \frac{7 + 4\sqrt{3}}{7 + 4\sqrt{3}} = \frac{7 + 4\sqrt{3}}{49 - (16 	imes 3)} = \frac{7 + 4\sqrt{3}}{49 - 48} = 7 + 4\sqrt{3}$.Now,find $x + \frac{1}{x}$:$x + \frac{1}{x} = (7 - 4\sqrt{3}) + (7 + 4\sqrt{3}) = 14$.We know that $(x + \frac{1}{x})^{2} = x^{2} + \frac{1}{x^{2}} + 2$.Therefore,$x^{2} + \frac{1}{x^{2}} = (x + \frac{1}{x})^{2} - 2$.Substituting the value:$x^{2} + \frac{1}{x^{2}} = (14)^{2} - 2 = 196 - 2 = 194$.

If $x = 7 - 4\sqrt{3}$,then find the value of $x^{2} + \frac{1}{x^{2}}$.

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