The value of 'k' for which one of the roots of `x^(2) - x + 3 k = 0`, is double of one of the roots of `x^(2) - x + k = 0`, is

Options

A. 1
B. -2
C. 2
D. none of these

Answer

To find the value of 'k' for which one of the roots of the equation \( x^2 - x + 3k = 0 \) is double of one of the roots of the equation \( x^2 - x + k = 0 \), we will follow these steps:

### Step 1: Define the roots
Let the roots of the equation \( x^2 - x + k = 0 \) be \( \alpha \) and \( \beta \). According to Vieta's formulas:
- Sum of the roots: \( \alpha + \beta = 1 \)
- Product of the roots: \( \alpha \beta = k \)

### Step 2: Express one root in terms of the other
Given that one root of \( x^2 - x + 3k = 0 \) is double one of the roots of \( x^2 - x + k = 0 \), we can assume:
- Let one root of \( x^2 - x + 3k = 0 \) be \( 2\alpha \).

### Step 3: Find the other root of the first equation
Let the other root of \( x^2 - x + 3k = 0 \) be \( \gamma \). Using Vieta's formulas again:
- Sum of the roots: \( 2\alpha + \gamma = 1 \)
- Product of the roots: \( 2\alpha \cdot \gamma = 3k \)

From the sum of the roots, we can express \( \gamma \) in terms of \( \alpha \):
\[ \gamma = 1 - 2\alpha \]

### Step 4: Substitute into the product of the roots
Substituting \( \gamma \) into the product of the roots equation:
\[ 2\alpha(1 - 2\alpha) = 3k \]
Expanding this gives:
\[ 2\alpha - 4\alpha^2 = 3k \]

### Step 5: Express \( k \) in terms of \( \alpha \)
Rearranging the equation gives:
\[ k = \frac{2\alpha - 4\alpha^2}{3} \]

### Step 6: Substitute \( k \) from the second equation
We know from the second equation that \( k = \alpha \beta \). From \( \alpha + \beta = 1 \), we can express \( \beta \) as:
\[ \beta = 1 - \alpha \]
Thus,
\[ k = \alpha(1 - \alpha) = \alpha - \alpha^2 \]

### Step 7: Equate the two expressions for \( k \)
Now we have two expressions for \( k \):
1. \( k = \frac{2\alpha - 4\alpha^2}{3} \)
2. \( k = \alpha - \alpha^2 \)

Setting them equal gives:
\[ \frac{2\alpha - 4\alpha^2}{3} = \alpha - \alpha^2 \]

### Step 8: Clear the fraction
Multiplying through by 3 to eliminate the fraction:
\[ 2\alpha - 4\alpha^2 = 3\alpha - 3\alpha^2 \]
Rearranging gives:
\[ -4\alpha^2 + 3\alpha^2 + 2\alpha - 3\alpha = 0 \]
This simplifies to:
\[ -\alpha^2 - \alpha = 0 \]
Factoring out \( -\alpha \):
\[ -\alpha(\alpha + 1) = 0 \]

### Step 9: Solve for \( \alpha \)
The solutions are:
\[ \alpha = 0 \quad \text{or} \quad \alpha = -1 \]

### Step 10: Find corresponding \( k \)
1. If \( \alpha = 0 \):
\[ k = 0(1 - 0) = 0 \]
2. If \( \alpha = -1 \):
\[ k = -1(1 - (-1)) = -1 \cdot 2 = -2 \]

Thus, the possible values of \( k \) are \( 0 \) and \( -2 \).

### Conclusion
The value of \( k \) for which one of the roots of \( x^2 - x + 3k = 0 \) is double of one of the roots of \( x^2 - x + k = 0 \) is:
\[ \boxed{-2} \]