If `alpha` and `beta` are the roots of the equation `x^2 +x+1 = 0`, the equation whose roots are `alpha^19` and `beta^7` 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`

Answer

To solve the problem, we need to find the quadratic equation whose roots are \( \alpha^{19} \) and \( \beta^{7} \), given that \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 + x + 1 = 0 \).

### Step-by-Step Solution:

1. **Identify the roots of the given equation**:
The roots \( \alpha \) and \( \beta \) can be found using the quadratic formula:
\[ \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 1, c = 1 \):
\[ \alpha, \beta = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]

2. **Calculate \( \alpha + \beta \) and \( \alpha \beta \)**:
From Vieta's formulas:
\[ \alpha + \beta = -\frac{b}{a} = -1 \]
\[ \alpha \beta = \frac{c}{a} = 1 \]

3. **Express \( \alpha^{19} \) and \( \beta^{7} \)**:
We know that \( \alpha^3 = 1 \) and \( \beta^3 = 1 \) (since \( \beta = \frac{1}{\alpha} \)).
Therefore:
\[ \alpha^{19} = \alpha^{18} \cdot \alpha = (\alpha^3)^6 \cdot \alpha = 1^6 \cdot \alpha = \alpha \]
\[ \beta^{7} = \beta^{6} \cdot \beta = (\beta^3)^2 \cdot \beta = 1^2 \cdot \beta = \beta \]

4. **Sum and product of the new roots**:
Now we find the sum and product of the new roots \( \alpha^{19} \) and \( \beta^{7} \):
\[ \text{Sum} = \alpha^{19} + \beta^{7} = \alpha + \beta = -1 \]
\[ \text{Product} = \alpha^{19} \cdot \beta^{7} = \alpha \cdot \beta = 1 \]

5. **Form the new quadratic equation**:
The quadratic equation with roots \( r_1 \) and \( r_2 \) can be expressed as:
\[ x^2 - (\text{Sum})x + (\text{Product}) = 0 \]
Substituting the values we found:
\[ x^2 - (-1)x + 1 = 0 \implies x^2 + x + 1 = 0 \]

### Final Answer:
The required quadratic equation is:
\[ \boxed{x^2 + x + 1 = 0} \]