If `omega` is an imaginary cube root of unity then the equation whose roots are `2omega + 3omega^(2)` and `2omega^(2) + 3omega` is

Options

A. a) `x^(2)+5x+7=0`
B. b) `x^(2)+5x-7=0`
C. c) `x^(2)-5x+7=0`
D. d) `x^(2)-5x-7=0`

Answer

To find the equation whose roots are \(2\omega + 3\omega^2\) and \(2\omega^2 + 3\omega\), where \(\omega\) is an imaginary cube root of unity, we will follow these steps:

### Step 1: Identify the Roots
The roots given are:
1. \( r_1 = 2\omega + 3\omega^2 \)
2. \( r_2 = 2\omega^2 + 3\omega \)

### Step 2: Calculate the Sum of the Roots
The sum of the roots \( S \) is given by:
\[ S = r_1 + r_2 = (2\omega + 3\omega^2) + (2\omega^2 + 3\omega) \]
Combining like terms:
\[ S = (2\omega + 3\omega) + (3\omega^2 + 2\omega^2) = 5\omega + 5\omega^2 = 5(\omega + \omega^2) \]

### Step 3: Use the Property of Cube Roots of Unity
Since \(\omega\) is a cube root of unity, we know:
\[ \omega + \omega^2 + 1 = 0 \implies \omega + \omega^2 = -1 \]
Thus, substituting this into our sum:
\[ S = 5(-1) = -5 \]

### Step 4: Calculate the Product of the Roots
The product of the roots \( P \) is given by:
\[ P = r_1 \cdot r_2 = (2\omega + 3\omega^2)(2\omega^2 + 3\omega) \]
Expanding this product:
\[ P = 2\omega \cdot 2\omega^2 + 2\omega \cdot 3\omega + 3\omega^2 \cdot 2\omega^2 + 3\omega^2 \cdot 3\omega \]
Calculating each term:
1. \( 2\omega \cdot 2\omega^2 = 4\omega^3 = 4(-1) = -4 \) (since \(\omega^3 = 1\))
2. \( 2\omega \cdot 3\omega = 6\omega^2 \)
3. \( 3\omega^2 \cdot 2\omega^2 = 6\omega^4 = 6\omega \) (since \(\omega^4 = \omega\))
4. \( 3\omega^2 \cdot 3\omega = 9\omega^3 = 9(-1) = -9 \)

Combining these results:
\[ P = -4 + 6\omega + 6\omega - 9 = -13 + 12\omega \]

### Step 5: Substitute \(\omega + \omega^2\) into the Product
Using \(\omega + \omega^2 = -1\):
\[ P = -13 + 12(-1) = -13 - 12 = -25 \]

### Step 6: Form the Quadratic Equation
The quadratic equation with roots \( r_1 \) and \( r_2 \) can be expressed as:
\[ x^2 - Sx + P = 0 \]
Substituting the values of \( S \) and \( P \):
\[ x^2 + 5x - 25 = 0 \]

### Final Equation
Thus, the equation whose roots are \(2\omega + 3\omega^2\) and \(2\omega^2 + 3\omega\) is:
\[ x^2 + 5x - 25 = 0 \]