If $$x + \frac{1}{x} = \frac{17}{4}, x > 1$$, then what is the value of $$x - \frac{1}{x}?$$
$$x+\frac{1}{x}=\frac{17}{4}$$
$$\left(x+\frac{1}{x}\right)^2=\frac{289}{16}$$
$$x^2+\frac{1}{x^2}+2=\frac{289}{16}$$
$$x^2+\frac{1}{x^2}=\frac{289}{16}-2$$
$$x^2+\frac{1}{x^2}=\frac{257}{16}$$
$$x^2+\frac{1}{x^2}-2=\frac{257}{16}-2$$
$$\left(x-\frac{1}{x}\right)^2=\frac{257-32}{16}$$
$$\left(x-\frac{1}{x}\right)^2=\frac{225}{16}$$
$$x-\frac{1}{x}=\frac{15}{4}$$
Hence, the correct answer is Option D
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