If `alpha + i beta` is one of the roots of the equation `x^(3)+qx + r=0`, then `2 alpha` is one of the roots of the equation :

Options

A. `x^(2)-qx+r=0`
B. `x^(3)-qx-r=0`
C. `x^(3)+qx-r=0`
D. none

Answer

To solve the problem, we need to determine the equation for which \( 2\alpha \) is a root, given that \( \alpha + i\beta \) is a root of the cubic equation \( x^3 + qx + r = 0 \).

### Step-by-Step Solution:

1. **Identify the roots of the cubic equation**:
Given that \( \alpha + i\beta \) is a root, its complex conjugate \( \alpha - i\beta \) is also a root of the equation. Let’s denote the third root as \( \gamma \).

2. **Use the property of the sum of roots**:
For a cubic equation \( x^3 + bx^2 + cx + d = 0 \), the sum of the roots is given by:
\[ \text{Sum of roots} = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} \]
In our case, since the coefficient of \( x^2 \) is 0 (the equation is \( x^3 + qx + r = 0 \)), we have:
\[ (\alpha + i\beta) + (\alpha - i\beta) + \gamma = 0 \]
Simplifying this gives:
\[ 2\alpha + \gamma = 0 \]
Therefore, we find:
\[ \gamma = -2\alpha \]

3. **Substituting the value of \( \gamma \)**:
Now we know that the roots of the equation are \( \alpha + i\beta \), \( \alpha - i\beta \), and \( -2\alpha \).

4. **Formulate the new equation**:
Since \( -2\alpha \) is a root, we can substitute \( x = -2\alpha \) into the original equation to find the new equation. We need to express the new equation in terms of \( x \).

5. **Substituting \( x = 2\alpha \)**:
To find the equation for which \( 2\alpha \) is a root, we can rewrite the original equation:
\[ (-2\alpha)^3 + q(-2\alpha) + r = 0 \]
This simplifies to:
\[ -8\alpha^3 - 2q\alpha + r = 0 \]
Multiplying through by -1 gives:
\[ 8\alpha^3 + 2q\alpha - r = 0 \]

6. **Rearranging the equation**:
We can express this in a standard polynomial form:
\[ x^3 + qx - r = 0 \]
where \( x = 2\alpha \).

### Conclusion:
The equation for which \( 2\alpha \) is a root is:
\[ x^3 + qx - r = 0 \]