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

$\frac{9}{4}$ $\frac{3}{2}$ $\frac{8}{3}$ $\frac{15}{4}$

$\frac{15}{4}$

If x + \(\frac{1}{x}\)  = n then  then, x - \(\frac{1}{x}\)  = \(\sqrt {n^2 - 4}\) If $x+\frac{1}{x}=\frac{17}{4}, x>1$ $x-\frac{1}{x}$ = \(\sqrt {(\frac{17}{4})^2 - 4}\) $x-\frac{1}{x}$ = \(\sqrt {\frac{225}{16}}\) = \(\frac{15}{4}\)