`alpha and beta ` are the roots of the quadratic equation `x^(2) - x - 1 = 0` what is the value of `alpha^(8) + beta^(8)` ?

Options

A. 47
B. 54
C. 59
D. 68

Answer

To find the value of \( \alpha^8 + \beta^8 \) where \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( x^2 - x - 1 = 0 \), we can follow these steps:

### Step 1: Identify the roots of the quadratic equation
The given quadratic equation is:
\[ x^2 - x - 1 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -1, c = -1 \):
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]
Thus, the roots are:
\[ \alpha = \frac{1 + \sqrt{5}}{2}, \quad \beta = \frac{1 - \sqrt{5}}{2} \]

### Step 2: Use the properties of roots
From Vieta's formulas, we know:
\[ \alpha + \beta = 1 \quad \text{and} \quad \alpha \beta = -1 \]

### Step 3: Find \( \alpha^2 + \beta^2 \)
Using the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the known values:
\[ \alpha^2 + \beta^2 = 1^2 - 2(-1) = 1 + 2 = 3 \]

### Step 4: Find \( \alpha^4 + \beta^4 \)
Using the identity:
\[ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2 \]
We already found \( \alpha^2 + \beta^2 = 3 \) and \( \alpha^2 \beta^2 = (\alpha \beta)^2 = (-1)^2 = 1 \):
\[ \alpha^4 + \beta^4 = 3^2 - 2 \cdot 1 = 9 - 2 = 7 \]

### Step 5: Find \( \alpha^8 + \beta^8 \)
Using the identity:
\[ \alpha^8 + \beta^8 = (\alpha^4 + \beta^4)^2 - 2\alpha^4\beta^4 \]
We found \( \alpha^4 + \beta^4 = 7 \) and \( \alpha^4 \beta^4 = (\alpha^2 \beta^2)^2 = 1^2 = 1 \):
\[ \alpha^8 + \beta^8 = 7^2 - 2 \cdot 1 = 49 - 2 = 47 \]

### Final Answer
Thus, the value of \( \alpha^8 + \beta^8 \) is:
\[ \boxed{47} \]