If $$ \frac{x}{y}+\frac{y}{x}=\frac{17}{4} $$, calculate two values of $$ \frac{x}{y} $$

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1. Let $$ t = \frac{x}{y} $$. Then $$\frac{y}{x} = \frac{1}{t}$$ and the equation becomes:
   $$
   t + \frac{1}{t} = \frac{17}{4}.
   $$

2. Multiply through by $$ t $$ to eliminate the fraction:
   $$
   t^2 + 1 = \frac{17}{4}t.
   $$

3. Rearrange the equation into standard quadratic form:
   $$
   t^2 - \frac{17}{4}t + 1 = 0.
   $$

4. Multiply the entire equation by $$ 4 $$ to clear the fraction:
   $$
   4t^2 - 17t + 4 = 0.
   $$

5. Use the quadratic formula:
   $$
   t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
   $$
   where $$ a = 4 $$, $$ b = -17 $$, and $$ c = 4 $$.

6. Substitute the values into the quadratic formula:
   $$
   t = \frac{17 \pm \sqrt{(-17)^2 - 4 \cdot 4 \cdot 4}}{2 \cdot 4}.
   $$

7. Calculate the discriminant:
   $$
   (-17)^2 - 4 \cdot 4 \cdot 4 = 289 - 64 = 225.
   $$

8. Simplify the square root:
   $$
   \sqrt{225} = 15.
   $$

9. Compute the two possible values of $$ t $$:
   $$
   t = \frac{17 + 15}{8} = \frac{32}{8} = 4,
   $$
   $$
   t = \frac{17 - 15}{8} = \frac{2}{8} = \frac{1}{4}.
   $$

Thus, the two values of $$\frac{x}{y}$$ are $$4$$ and $$\frac{1}{4}$$.
The two values of $$\frac{x}{y}$$ are 4 and $$\frac{1}{4}$$.

If , calculate two values of

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