If `alpha and beta` are roots of the equation`x^2 – x-1 = 0`, then the equation where roots are `alpha/beta and beta/alpha` is:

Options

A. `x^2 + 3x – 1= 0`
B. `x^2 + x-1=0`
C. `x^2 – x+1 = 0`
D. `x^2 + 3x + 1 = 0`

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Identify the roots of the given equation
The given equation is:
\[ x^2 - x - 1 = 0 \]

### Step 2: Calculate the sum and product of the roots
Using Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = 1 \)
- The product of the roots \( \alpha \beta = \frac{c}{a} = -1 \)

### Step 3: Find the sum of the new roots \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \)
We can express this as:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} \]

### Step 4: Calculate \( \alpha^2 + \beta^2 \)
Using the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \]
Substituting the values we found:
\[ \alpha^2 + \beta^2 = (1)^2 - 2(-1) = 1 + 2 = 3 \]

### Step 5: Substitute back to find the sum of the new roots
Now substituting back:
\[ \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{3}{-1} = -3 \]
Thus, the sum of the new roots is:
\[ S = -3 \]

### Step 6: Find the product of the new roots \( \frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} \)
This simplifies to:
\[ \frac{\alpha \beta}{\beta \alpha} = 1 \]
Thus, the product of the new roots is:
\[ P = 1 \]

### Step 7: Form the new quadratic equation
Using the standard form of a quadratic equation:
\[ x^2 - Sx + P = 0 \]
Substituting \( S \) and \( P \):
\[ x^2 - (-3)x + 1 = 0 \]
This simplifies to:
\[ x^2 + 3x + 1 = 0 \]

### Final Answer
The equation where the roots are \( \frac{\alpha}{\beta} \) and \( \frac{\beta}{\alpha} \) is:
\[ x^2 + 3x + 1 = 0 \]

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