If p and q are rational numbers and $\frac{3+\sqrt{5}}{3-\sqrt{5}}=p+q\sqrt{5}$, then value of p and q are :

Question

Accepted Answer

To solve the problem and verify why Option 3 is the correct answer, we need to simplify the expression $\frac{3+\sqrt{5}}{3-\sqrt{5}}$ and express it in the form $p + q\sqrt{5}$, where $p$ and $q$ are rational numbers.

### Step-by-Step Solution:

1. **Rationalize the Denominator:**
   To simplify $\frac{3+\sqrt{5}}{3-\sqrt{5}}$, we multiply the numerator and the denominator by the conjugate of the denominator, which is $3+\sqrt{5}$.

$$
   \frac{3+\sqrt{5}}{3-\sqrt{5}} \times \frac{3+\sqrt{5}}{3+\sqrt{5}} = \frac{(3+\sqrt{5})^2}{(3-\sqrt{5})(3+\sqrt{5})}
   $$

2. **Simplify the Denominator:**
   The denominator becomes a difference of squares:

$$
   (3-\sqrt{5})(3+\sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4
   $$

3. **Expand the Numerator:**
   Expand $(3+\sqrt{5})^2$:

$$
   (3+\sqrt{5})^2 = 3^2 + 2 \cdot 3 \cdot \sqrt{5} + (\sqrt{5})^2 = 9 + 6\sqrt{5} + 5 = 14 + 6\sqrt{5}
   $$

4. **Combine the Results:**
   Now, substitute back into the fraction:

$$
   \frac{14 + 6\sqrt{5}}{4} = \frac{14}{4} + \frac{6\sqrt{5}}{4} = \frac{7}{2} + \frac{3}{2}\sqrt{5}
   $$

5. **Identify $p$ and $q$:**
   From the expression $\frac{7}{2} + \frac{3}{2}\sqrt{5}$, we can identify:
   - $p = \frac{7}{2}$
   - $q = \frac{3}{2}$

### Conclusion:

- **Option 3**: $p = \frac{7}{2}$ and $q = \frac{3}{2}$ is indeed the correct answer as derived from the simplification process.

### Explanation of Incorrect Options:

- **Option 1**: $p = \frac{2}{7}$; $q = \frac{3}{2}$ is incorrect because the value of $p$ does not match the derived $\frac{7}{2}$.
- **Option 2**: $p = \frac{7}{2}$; $q = \frac{2}{7}$ is incorrect because the value of $q$ does not match the derived $\frac{3}{2}$.
- **Option 4**: $p = \frac{7}{2}$; $q = \frac{7}{2}$ is incorrect because the value of $q$ does not match the derived $\frac{3}{2}$.

Thus, the correct answer is Option 3, where $p = \frac{7}{2}$ and $q = \frac{3}{2}$.