Sum of sixth power of the roots of the equation `t^(2) - 2t + 4 = 0` is

Options

A. 256
B. 128
C. 64
D. 32

Answer

To find the sum of the sixth power of the roots of the equation \( t^2 - 2t + 4 = 0 \), we can follow these steps:

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

Calculating the discriminant:
\[ b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot 4 = 4 - 16 = -12 \]
Since the discriminant is negative, the roots will be complex.

Now substituting into the quadratic formula:
\[ t = \frac{2 \pm \sqrt{-12}}{2} = \frac{2 \pm 2i\sqrt{3}}{2} = 1 \pm i\sqrt{3} \]

Thus, the roots are:
\[ \alpha = 1 + i\sqrt{3}, \quad \beta = 1 - i\sqrt{3} \]

### Step 2: Calculate the sixth power of the roots
We need to find \( \alpha^6 + \beta^6 \).

First, we express \( \alpha \) and \( \beta \) in polar form:
\[ \alpha = 2 \left( \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} \right), \quad \beta = 2 \left( \cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right) \right) \]

### Step 3: Use De Moivre's Theorem
Using De Moivre's theorem, we can calculate the sixth powers:
\[ \alpha^6 = \left(2 \text{cis} \frac{\pi}{3}\right)^6 = 2^6 \text{cis}(2\pi) = 64(\cos 2\pi + i\sin 2\pi) = 64(1 + 0i) = 64 \]
\[ \beta^6 = \left(2 \text{cis} \left(-\frac{\pi}{3}\right)\right)^6 = 2^6 \text{cis}(-2\pi) = 64(\cos(-2\pi) + i\sin(-2\pi)) = 64(1 + 0i) = 64 \]

### Step 4: Sum the sixth powers
Now, we can sum the sixth powers:
\[ \alpha^6 + \beta^6 = 64 + 64 = 128 \]

### Final Answer
Thus, the sum of the sixth power of the roots is:
\[ \boxed{128} \]