If x = sqrt{frac{2+sqrt{3}}{2-sqrt{3}}},then | vedclass

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(NONE OF THE ABOVE (CALCULATED VALUE IS $-4 + 3\sqrt{3}$)) Given $x = \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}$.
Rationalizing the denominator inside the

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$ -4 + 3\sqrt{3}$

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(NONE OF THE ABOVE (CALCULATED VALUE IS $-4 + 3\sqrt{3}$)) Given $x = \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}$.
Rationalizing the denominator inside the square root:
$x = \sqrt{\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}} = \sqrt{\frac{(2+\sqrt{3})^2}{4-3}} = \sqrt{(2+\sqrt{3})^2} = 2+\sqrt{3}$.
Now,we need to find the value of $(x^2 - x - 9)$.
Substitute $x = 2+\sqrt{3}$ into the expression:
$x^2 = (2+\sqrt{3})^2 = 4 + 3 + 4\sqrt{3} = 7 + 4\sqrt{3}$.
Now,$(x^2 - x - 9) = (7 + 4\sqrt{3}) - (2 + \sqrt{3}) - 9$.
$= 7 + 4\sqrt{3} - 2 - \sqrt{3} - 9$.
$= (7 - 2 - 9) + (4\sqrt{3} - \sqrt{3})$.
$= -4 + 3\sqrt{3}$.

If $x = \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}$,then what is the value of $(x^{2} - x - 9)$?

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