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Question:

If x + $\frac{1}{x}$ = $\frac{17}{4}$, x > 1, then what is the value of x - $\frac{1}{x}$?

Explanation:

x + $\frac{1}{x}$ = $\frac{17}{4}$

${\left(x+\frac{1}{x}\right)}^2$ = $\frac{289}{16}$

x² + $\frac{1}{x^2}$ + 2 = $\frac{289}{16}$

x² + $\frac{1}{x^2}$ = $\frac{289}{16}$ - 2

x² + $\frac{1}{x^2}$ = $\frac{257}{16}$

x² + $\frac{1}{x^2}$ - 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