CUET Preparation Today
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 $x+\frac{1}{x}=\frac{17}{4}, x>1$ $x-\frac{1}{x}$ = \(\sqrt {(\frac{17}{4})^2 - 4}\) $x-\frac{1}{x}$ = \(\sqrt {\frac{225}{16}}\) = \(\frac{15}{4}\) |
