If $\alpha, \beta$ are the roots of $x^{2}+x+1=0$ and $\frac{\alpha}{\beta},\frac{\beta}{\alpha}$ are the roots of $x^{2}+px+q=0$ , then the value of p is:

Question

Accepted Answer

To solve this problem, we need to find the value of $ p $ in the quadratic equation $ x^2 + px + q = 0 $ given that the roots are $ \frac{\alpha}{\beta} $ and $ \frac{\beta}{\alpha} $, where $ \alpha $ and $ \beta $ are the roots of the equation $ x^2 + x + 1 = 0 $.

### Step-by-step Explanation:

1. **Find the Roots $\alpha$ and $\beta$:**

The roots of the equation $ x^2 + x + 1 = 0 $ can be found using the quadratic formula:
   $$
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   $$
   Here, $ a = 1 $, $ b = 1 $, and $ c = 1 $. Plugging these values into the formula gives:
   $$
   x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2}
   $$
   $$
   x = \frac{-1 \pm i\sqrt{3}}{2}
   $$
   Thus, the roots are $ \alpha = \frac{-1 + i\sqrt{3}}{2} $ and $ \beta = \frac{-1 - i\sqrt{3}}{2} $.

2. **Calculate $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$:**

$$
   \frac{\alpha}{\beta} = \frac{\frac{-1 + i\sqrt{3}}{2}}{\frac{-1 - i\sqrt{3}}{2}} = \frac{-1 + i\sqrt{3}}{-1 - i\sqrt{3}}
   $$
   Multiply numerator and denominator by the conjugate of the denominator:
   $$
   \frac{\alpha}{\beta} = \frac{(-1 + i\sqrt{3})(-1 + i\sqrt{3})}{(-1 - i\sqrt{3})(-1 + i\sqrt{3})} = \frac{1 - 2i\sqrt{3} - 3}{1 + 3} = \frac{-2 - 2i\sqrt{3}}{4} = -\frac{1}{2} - \frac{i\sqrt{3}}{2}
   $$

Similarly, calculate $\frac{\beta}{\alpha}$:
   $$
   \frac{\beta}{\alpha} = \frac{-1 - i\sqrt{3}}{-1 + i\sqrt{3}} = \frac{(-1 - i\sqrt{3})(-1 - i\sqrt{3})}{(-1 + i\sqrt{3})(-1 - i\sqrt{3})} = \frac{1 + 2i\sqrt{3} - 3}{1 + 3} = \frac{-2 + 2i\sqrt{3}}{4} = -\frac{1}{2} + \frac{i\sqrt{3}}{2}
   $$

3. **Sum of the Roots:**

The sum of the roots $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$ is:
   $$
   \left(-\frac{1}{2} - \frac{i\sqrt{3}}{2}\right) + \left(-\frac{1}{2} + \frac{i\sqrt{3}}{2}\right) = -1
   $$

According to Vieta's formulas, the sum of the roots of the quadratic equation $ x^2 + px + q = 0 $ is equal to $-p$. Therefore:
   $$
   -p = -1 \implies p = 1
   $$

### Conclusion:

- **Correct Option:** The value of $ p $ is 1, which corresponds to Option 3.
- **Incorrect Options:**
  - Option 1 (-2), Option 2 (-1), and Option 4 (2) are incorrect because they do not match the calculated value of $ p $ based on the sum of the roots derived from Vieta's formulas.

Thus, the correct answer is indeed Option 3, $ p = 1 $.