The quadratic equations `x^(2)-6x+a=0 and x^(2)-cx+6=0` have one root in common. The other roots of the first and second equations are integers in the ratio 4:3. then the common root is

Options

A. 1
B. 4
C. 3
D. 2

Answer

To solve the problem, we need to find the common root of the two quadratic equations given that they share one root and the other roots are in the ratio of 4:3. Let's go through the solution step by step.

### Step 1: Define the common root and other roots
Let the common root be denoted as \( \alpha \). For the first equation \( x^2 - 6x + a = 0 \), let the other root be \( \beta_1 \). For the second equation \( x^2 - cx + 6 = 0 \), let the other root be \( \beta_2 \). According to the problem, the ratio of the other roots is given as:
\[ \frac{\beta_1}{\beta_2} = \frac{4}{3} \]
We can express the roots as:
\[ \beta_1 = 4k \quad \text{and} \quad \beta_2 = 3k \]
for some integer \( k \).

### Step 2: Use the product of roots for the first equation
From Vieta's formulas, we know that the product of the roots of the first equation is equal to \( a \):
\[ \alpha \cdot \beta_1 = a \]
Substituting \( \beta_1 \):
\[ \alpha \cdot 4k = a \quad \text{(1)} \]

### Step 3: Use the product of roots for the second equation
Similarly, for the second equation, the product of the roots is equal to 6:
\[ \alpha \cdot \beta_2 = 6 \]
Substituting \( \beta_2 \):
\[ \alpha \cdot 3k = 6 \quad \text{(2)} \]

### Step 4: Solve for \( \alpha \) and \( k \)
From equation (2), we can express \( k \) in terms of \( \alpha \):
\[ 3k = \frac{6}{\alpha} \quad \Rightarrow \quad k = \frac{2}{\alpha} \]

### Step 5: Substitute \( k \) into equation (1)
Now substitute \( k \) back into equation (1):
\[ \alpha \cdot 4\left(\frac{2}{\alpha}\right) = a \]
This simplifies to:
\[ \frac{8}{\alpha} = a \]

### Step 6: Find the value of \( a \)
Now we can express \( a \) in terms of \( \alpha \):
\[ a = \frac{8}{\alpha} \]

### Step 7: Find integer values for \( \beta_1 \) and \( \beta_2 \)
Since \( \beta_1 = 4k = 4\left(\frac{2}{\alpha}\right) = \frac{8}{\alpha} \) and \( \beta_2 = 3k = 3\left(\frac{2}{\alpha}\right) = \frac{6}{\alpha} \), we need both \( \beta_1 \) and \( \beta_2 \) to be integers. This implies that \( \alpha \) must be a divisor of both 8 and 6.

### Step 8: Determine possible values for \( \alpha \)
The common divisors of 8 and 6 are 1 and 2. Thus, the possible values for \( \alpha \) are 1 and 2.

### Step 9: Check which value satisfies the integer condition
1. If \( \alpha = 2 \):
- \( k = \frac{2}{2} = 1 \)
- \( \beta_1 = 4k = 4 \)
- \( \beta_2 = 3k = 3 \)
- Both are integers.

2. If \( \alpha = 1 \):
- \( k = \frac{2}{1} = 2 \)
- \( \beta_1 = 4k = 8 \)
- \( \beta_2 = 3k = 6 \)
- Both are integers.

### Conclusion
However, since we are looking for the common root, we find that \( \alpha = 2 \) is the only value that maintains the integer condition for the roots of both equations.

Thus, the common root is:
\[ \boxed{2} \]