The quadratic equations `x^2""-6x""+""a""=""0""a n d ""x^2""-c x""+""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. 4
B. 3
C. 2
D. 1

Explanation

To solve the problem, we need to find the common root of the two quadratic equations given that the other roots are in the ratio of 4:3.

### Step-by-step Solution:

1. **Identify the Quadratic Equations:**
The two quadratic equations are:
- \( x^2 - 6x + a = 0 \) (Equation 1)
- \( x^2 - cx + 6 = 0 \) (Equation 2)

2. **Let the Common Root be \( \alpha \):**
Let \( \alpha \) be the common root of both equations. The other root of Equation 1 can be denoted as \( 4\beta \) and the other root of Equation 2 can be denoted as \( 3\beta \).

3. **Use Vieta's Formulas:**
From Equation 1:
- Sum of roots: \( \alpha + 4\beta = 6 \) (1)
- Product of roots: \( \alpha \cdot 4\beta = a \) (2)

From Equation 2:
- Sum of roots: \( \alpha + 3\beta = c \) (3)
- Product of roots: \( \alpha \cdot 3\beta = 6 \) (4)

4. **Express \( \beta \) in terms of \( \alpha \):**
From Equation (4):
\[ \alpha \cdot 3\beta = 6 \implies \beta = \frac{2}{\alpha} \]

5. **Substitute \( \beta \) into Equation (1):**
Substitute \( \beta \) into Equation (1):
\[ \alpha + 4\left(\frac{2}{\alpha}\right) = 6 \]
This simplifies to:
\[ \alpha + \frac{8}{\alpha} = 6 \]

6. **Multiply through by \( \alpha \) to eliminate the fraction:**
\[ \alpha^2 + 8 = 6\alpha \]
Rearranging gives:
\[ \alpha^2 - 6\alpha + 8 = 0 \]

7. **Solve the Quadratic Equation:**
We can use the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -6, c = 8 \):
\[ \alpha = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} \]
\[ = \frac{6 \pm \sqrt{36 - 32}}{2} \]
\[ = \frac{6 \pm \sqrt{4}}{2} \]
\[ = \frac{6 \pm 2}{2} \]
Thus, we have:
\[ \alpha = \frac{8}{2} = 4 \quad \text{or} \quad \alpha = \frac{4}{2} = 2 \]

8. **Conclusion:**
The common root \( \alpha \) can be either \( 2 \) or \( 4 \).