If `alpha, beta` are roots of the equation `x^(2) + x + 1 = 0`, then the equation whose roots are `(alpha)/(beta) and (beta)/(alpha)`, is

Options

A. `x^(2) + x + 1 = 0`
B. `x^(2) - x + 1 = 0`
C. `x^(2) - x - 1 = 0`
D. `x^(2) + x - 1 = 0`

Explanation

To find the equation whose roots are \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\), where \(\alpha\) and \(\beta\) are the roots of the equation \(x^2 + x + 1 = 0\), we can follow these steps:

### Step 1: Determine 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 \(c = 1\).
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]
Thus, the roots are:
\[ \alpha = \frac{-1 + i\sqrt{3}}{2}, \quad \beta = \frac{-1 - i\sqrt{3}}{2} \]

### Step 2: Calculate the sum and product of the new roots
The new roots are \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\).

**Sum of the new roots:**
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} \]
We know that \(\alpha + \beta = -1\) and \(\alpha \beta = 1\). To find \(\alpha^2 + \beta^2\):
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = (-1)^2 - 2 \cdot 1 = 1 - 2 = -1 \]
Thus, the sum of the new roots is:
\[ \frac{-1}{1} = -1 \]

**Product of the new roots:**
\[ \frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} = 1 \]

### Step 3: Form the new quadratic equation
Using the sum and product of the roots, we can form the quadratic equation:
\[ x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \]
Substituting the values we found:
\[ x^2 - (-1)x + 1 = 0 \implies x^2 + x + 1 = 0 \]

### Final Answer
The required equation whose roots are \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\) is:
\[ \boxed{x^2 + x + 1 = 0} \]