Problem - 3324
If $x^2 + 11x+16=0, y^2 + 11y+16=0$, and $x\ne y$, what is the value of $$\sqrt{\frac{x}{y}}-\sqrt{\frac{y}{x}}$$
Clearly, $x$ and $y$ are the two roots of equation $t^2 + 11t+16=0$. Hence
$$\sqrt{\frac{x}{y}}-\sqrt{\frac{y}{x}}=\frac{y-x}{\sqrt{xy}}=\frac{\pm\sqrt{(x+y)^2-4xy}}{\sqrt{xy}}=\frac{\pm\sqrt{11^2-4\times 16}}{\sqrt{16}}=\boxed{\pm\frac{\sqrt{57}}{4}}$$