If \(x = \sqrt{\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}}\), determine the value of \(x^2 + x - 9\).
5√3
First simplify \(x^2 = \dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}\).
Rationalise: multiply numerator and denominator by \(2 + \sqrt{3}\):
\(x^2 = \dfrac{(2 + \sqrt{3})^2}{(2 - \sqrt{3})(2 + \sqrt{3})} = \dfrac{4 + 4\sqrt{3} + 3}{4 - 3} = \dfrac{7 + 4\sqrt{3}}{1} = 7 + 4\sqrt{3}\).
Now find x: note that \((2 + \sqrt{3})^2 = 7 + 4\sqrt{3}\), so \(x = 2 + \sqrt{3}\).
Compute \(x^2 + x - 9 = (7 + 4\sqrt{3}) + (2 + \sqrt{3}) - 9 = 0 + 5\sqrt{3} = 5\sqrt{3}\).
Hence, the answer is 5√3.
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