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 the problem of finding \( \alpha^{15} + \beta^{15} \) where \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 + 2x + 2 = 0 \), we will follow these steps:

### Step 1: Find the roots \( \alpha \) and \( \beta \)

The roots of the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our equation \( x^2 + 2x + 2 = 0 \), we have \( a = 1, b = 2, c = 2 \).

Substituting these values into the formula:

\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2} \]

This simplifies to:

\[ x = \frac{-2 \pm 2i}{2} = -1 \pm i \]

Thus, the roots are:

\[ \alpha = -1 + i \quad \text{and} \quad \beta = -1 - i \]

### Step 2: Calculate \( \alpha^2 \) and \( \beta^2 \)

Now we will compute \( \alpha^2 \) and \( \beta^2 \):

\[ \alpha^2 = (-1 + i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i \]

\[ \beta^2 = (-1 - i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i \]

### Step 3: Use the recurrence relation

We can use the relation \( \alpha^n + \beta^n = -2(\alpha^{n-1} + \beta^{n-1}) - (\alpha^{n-2} + \beta^{n-2}) \).

Let \( S_n = \alpha^n + \beta^n \). We have:

- \( S_0 = 2 \) (since \( \alpha^0 + \beta^0 = 1 + 1 \))
- \( S_1 = -2 \) (since \( \alpha + \beta = -2 \))

Now, using the recurrence relation:

\[ S_2 = \alpha^2 + \beta^2 = -2 \]
\[ S_3 = -2S_2 - S_1 = -2(-2) - (-2) = 4 + 2 = 6 \]
\[ S_4 = -2S_3 - S_2 = -2(6) - (-2) = -12 + 2 = -10 \]
\[ S_5 = -2S_4 - S_3 = -2(-10) - 6 = 20 - 6 = 14 \]
\[ S_6 = -2S_5 - S_4 = -2(14) - (-10) = -28 + 10 = -18 \]
\[ S_7 = -2S_6 - S_5 = -2(-18) - 14 = 36 - 14 = 22 \]
\[ S_8 = -2S_7 - S_6 = -2(22) - (-18) = -44 + 18 = -26 \]
\[ S_9 = -2S_8 - S_7 = -2(-26) - 22 = 52 - 22 = 30 \]
\[ S_{10} = -2S_9 - S_8 = -2(30) - (-26) = -60 + 26 = -34 \]
\[ S_{11} = -2S_{10} - S_9 = -2(-34) - 30 = 68 - 30 = 38 \]
\[ S_{12} = -2S_{11} - S_{10} = -2(38) - (-34) = -76 + 34 = -42 \]
\[ S_{13} = -2S_{12} - S_{11} = -2(-42) - 38 = 84 - 38 = 46 \]
\[ S_{14} = -2S_{13} - S_{12} = -2(46) - (-42) = -92 + 42 = -50 \]
\[ S_{15} = -2S_{14} - S_{13} = -2(-50) - 46 = 100 - 46 = 54 \]

### Final Result

Thus, \( \alpha^{15} + \beta^{15} = S_{15} = 54 \).