If `alpha and beta` are the roots of `x^(2)-x-2=0`, find the quadratic equation in x whose roots are `(2alpha+1)` and `(2beta+1):`

Options

A. `x^(2)+5x-4=0`
B. `x^(2)-4x-5=0`
C. `x^(2)+4x-5=0`
D. none of these

Answer

To solve the problem, we need to find the quadratic equation whose roots are \(2\alpha + 1\) and \(2\beta + 1\), given that \(\alpha\) and \(\beta\) are the roots of the equation \(x^2 - x - 2 = 0\).

### Step-by-Step Solution:

1. **Find the roots \(\alpha\) and \(\beta\)**:
We start with the quadratic equation:
\[ x^2 - x - 2 = 0 \]
We can factor this equation:
\[ (x - 2)(x + 1) = 0 \]
Thus, the roots are:
\[ \alpha = 2 \quad \text{and} \quad \beta = -1 \]

2. **Calculate \(2\alpha + 1\) and \(2\beta + 1\)**:
Now we compute the new roots:
\[ 2\alpha + 1 = 2(2) + 1 = 4 + 1 = 5 \]
\[ 2\beta + 1 = 2(-1) + 1 = -2 + 1 = -1 \]

3. **Find the sum and product of the new roots**:
The sum of the new roots \( (2\alpha + 1) + (2\beta + 1) \) is:
\[ 5 + (-1) = 4 \]
The product of the new roots \( (2\alpha + 1)(2\beta + 1) \) is:
\[ 5 \times (-1) = -5 \]

4. **Form the new quadratic equation**:
The general form of a quadratic equation with roots \(r_1\) and \(r_2\) is given by:
\[ x^2 - (r_1 + r_2)x + (r_1 \cdot r_2) = 0 \]
Substituting the sum and product of the new roots:
\[ x^2 - (4)x + (-5) = 0 \]
This simplifies to:
\[ x^2 - 4x - 5 = 0 \]

### Final Answer:
The quadratic equation whose roots are \(2\alpha + 1\) and \(2\beta + 1\) is:
\[ \boxed{x^2 - 4x - 5 = 0} \]