Let `alpha` and `beta` be two roots of the equation `x^(2) + 2x + 2 = 0`. Then `alpha^(15) + beta^(15)` is equal to

Options

A. 512
B. -512
C. -256
D. 256

Answer

To solve for \( \alpha^{15} + \beta^{15} \) where \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 + 2x + 2 = 0 \), we can follow these steps:

### Step 1: Find the roots of the quadratic equation
The roots of the quadratic equation \( x^2 + 2x + 2 = 0 \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 2 \), and \( c = 2 \).

Calculating the discriminant:
\[ b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 2 = 4 - 8 = -4 \]
Since the discriminant is negative, the roots are complex.

Now substituting into the quadratic formula:
\[ x = \frac{-2 \pm \sqrt{-4}}{2 \cdot 1} = \frac{-2 \pm 2i}{2} = -1 \pm i \]
Thus, the roots are:
\[ \alpha = -1 - i \quad \text{and} \quad \beta = -1 + i \]

### Step 2: Use the properties of roots
We can express \( \alpha^{15} + \beta^{15} \) using the recurrence relation derived from the roots. For any quadratic equation \( x^2 + bx + c = 0 \), the roots satisfy:
\[ \alpha^n + \beta^n = -b(\alpha^{n-1} + \beta^{n-1}) - c(\alpha^{n-2} + \beta^{n-2}) \]
For our equation, \( b = 2 \) and \( c = 2 \). Thus, we have:
\[ \alpha^n + \beta^n = -2(\alpha^{n-1} + \beta^{n-1}) - 2(\alpha^{n-2} + \beta^{n-2}) \]

### Step 3: Calculate initial values
We need to calculate \( \alpha^0 + \beta^0 \) and \( \alpha^1 + \beta^1 \):
\[ \alpha^0 + \beta^0 = 2 \]
\[ \alpha^1 + \beta^1 = (-1 - i) + (-1 + i) = -2 \]

### Step 4: Use the recurrence relation
Now, we can calculate the next values using the recurrence relation:
\[ \alpha^2 + \beta^2 = -2(-2) - 2(2) = 4 - 4 = 0 \]
\[ \alpha^3 + \beta^3 = -2(0) - 2(-2) = 0 + 4 = 4 \]
\[ \alpha^4 + \beta^4 = -2(4) - 2(0) = -8 \]
\[ \alpha^5 + \beta^5 = -2(-8) - 2(4) = 16 - 8 = 8 \]
\[ \alpha^6 + \beta^6 = -2(8) - 2(-8) = -16 + 16 = 0 \]
\[ \alpha^7 + \beta^7 = -2(0) - 2(8) = -16 \]
\[ \alpha^8 + \beta^8 = -2(-16) - 2(0) = 32 \]
\[ \alpha^9 + \beta^9 = -2(32) - 2(-16) = -64 + 32 = -32 \]
\[ \alpha^{10} + \beta^{10} = -2(-32) - 2(32) = 64 - 64 = 0 \]
\[ \alpha^{11} + \beta^{11} = -2(0) - 2(-32) = 64 \]
\[ \alpha^{12} + \beta^{12} = -2(64) - 2(0) = -128 \]
\[ \alpha^{13} + \beta^{13} = -2(-128) - 2(64) = 256 - 128 = 128 \]
\[ \alpha^{14} + \beta^{14} = -2(128) - 2(-128) = -256 + 256 = 0 \]
\[ \alpha^{15} + \beta^{15} = -2(0) - 2(128) = -256 \]

### Final Answer
Thus, the value of \( \alpha^{15} + \beta^{15} \) is:
\[ \alpha^{15} + \beta^{15} = -256 \]