If `cos^4theta+alpha` are the roots of the equation `x^2+2b x+b=0a n dcos^2theta+beta,sin^2theta+betaa r e` the roots of the equation `x^2+4x+2=0,` then values of `b` are `2` b. `-1` c. `-2` d. `2`

Options

A. 2
B. -1
C. -2
D. 1

Explanation

To solve the problem, we need to analyze the given equations and their roots step by step.

### Step 1: Identify the roots of the first equation
The first equation is given as:
\[ x^2 + 2bx + b = 0 \]
Let the roots of this equation be:
\[ r_1 = \cos^4 \theta + \alpha \]
\[ r_2 = \sin^4 \theta + \alpha \]

### Step 2: Use the sum and product of roots
From Vieta's formulas, we know:
- The sum of the roots \( r_1 + r_2 = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -2b \)
- The product of the roots \( r_1 \cdot r_2 = \frac{\text{constant term}}{\text{coefficient of } x^2} = b \)

Thus, we have:
\[ r_1 + r_2 = \cos^4 \theta + \sin^4 \theta + 2\alpha = -2b \]
\[ r_1 \cdot r_2 = (\cos^4 \theta + \alpha)(\sin^4 \theta + \alpha) = b \]

### Step 3: Identify the roots of the second equation
The second equation is given as:
\[ x^2 + 4x + 2 = 0 \]
Let the roots of this equation be:
\[ s_1 = \cos^2 \theta + \beta \]
\[ s_2 = \sin^2 \theta + \beta \]

### Step 4: Use the sum and product of roots for the second equation
From Vieta's formulas for the second equation:
- The sum of the roots \( s_1 + s_2 = -4 \)
- The product of the roots \( s_1 \cdot s_2 = 2 \)

Thus, we have:
\[ s_1 + s_2 = \cos^2 \theta + \sin^2 \theta + 2\beta = -4 \]
Since \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ 1 + 2\beta = -4 \]
\[ 2\beta = -5 \]
\[ \beta = -\frac{5}{2} \]

### Step 5: Relate the roots of the first and second equations
We know:
\[ \cos^4 \theta - \sin^4 \theta = (\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta) \]
Since \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ \cos^4 \theta - \sin^4 \theta = \cos^2 \theta - \sin^2 \theta \]

### Step 6: Calculate the difference of roots
The difference of roots for the first equation can be calculated as:
\[ \sqrt{(r_1 + r_2)^2 - 4r_1r_2} = \sqrt{(-2b)^2 - 4b} = \sqrt{4b^2 - 4b} = 2\sqrt{b(b-1)} \]

For the second equation:
\[ \sqrt{(s_1 + s_2)^2 - 4s_1s_2} = \sqrt{(-4)^2 - 4 \cdot 2} = \sqrt{16 - 8} = \sqrt{8} = 2\sqrt{2} \]

### Step 7: Equate the differences
Setting the two differences equal gives:
\[ 2\sqrt{b(b-1)} = 2\sqrt{2} \]
Dividing both sides by 2:
\[ \sqrt{b(b-1)} = \sqrt{2} \]
Squaring both sides:
\[ b(b-1) = 2 \]
\[ b^2 - b - 2 = 0 \]

### Step 8: Factor the quadratic equation
Factoring gives:
\[ (b - 2)(b + 1) = 0 \]
Thus, the solutions are:
\[ b = 2 \quad \text{or} \quad b = -1 \]

### Final Answer
The values of \( b \) are:
\[ b = 2 \quad \text{and} \quad b = -1 \]