Calculate $\frac{1}{x}$
Given the value of $x$ is $2 + \sqrt{3}$. To find $\frac{1}{x}$, we calculate the reciprocal:
$ \frac{1}{x} = \frac{1}{2 + \sqrt{3}} $To simplify this fraction, we multiply the numerator and the denominator by the conjugate of the denominator, which is $2 - \sqrt{3}$:
$ \frac{1}{x} = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} $Using the difference of squares formula $(a+b)(a-b) = a^2 - b^2$ for the denominator:
$ \frac{1}{x} = \frac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2} = \frac{2 - \sqrt{3}}{4 - 3} = \frac{2 - \sqrt{3}}{1} $Therefore, $\frac{1}{x} = 2 - \sqrt{3}$.
Calculate $x - \frac{1}{x}$
Now, substitute the values of $x$ and $\frac{1}{x}$ into the expression $x - \frac{1}{x}$:
$ x - \frac{1}{x} = (2 + \sqrt{3}) - (2 - \sqrt{3}) $Distribute the negative sign:
$ x - \frac{1}{x} = 2 + \sqrt{3} - 2 + \sqrt{3} $Combine like terms:
$ x - \frac{1}{x} = (2 - 2) + (\sqrt{3} + \sqrt{3}) $ $ x - \frac{1}{x} = 0 + 2\sqrt{3} $Thus, the final value is $2\sqrt{3}$.
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