If `a,b,c, in R` and equations `ax^(2) + bx + c =0` and `x^(2) + 2x + 9 = 0` have a common root then

Options

A. `a:b:c=1:2:9`
B. `a:b:c=9:2:4`
C. `a:b:c=9:2:1`
D. `a:b:c=2:1:9`

Answer

To solve the problem, we need to find the relationship between the coefficients \(a\), \(b\), and \(c\) given that the equations \(ax^2 + bx + c = 0\) and \(x^2 + 2x + 9 = 0\) have a common root.

### Step-by-Step Solution:

1. **Identify the equations:**
- Let \(f(x) = ax^2 + bx + c = 0\)
- Let \(g(x) = x^2 + 2x + 9 = 0\)

2. **Assume a common root:**
- Let \(\alpha\) be the common root of both equations. Therefore, we have:
\[ f(\alpha) = a\alpha^2 + b\alpha + c = 0 \]
\[ g(\alpha) = \alpha^2 + 2\alpha + 9 = 0 \]

3. **Express \(g(\alpha) = 0\):**
- From \(g(\alpha) = 0\), we can express:
\[ \alpha^2 + 2\alpha + 9 = 0 \quad \text{(1)} \]

4. **Substitute \(\alpha^2\) from equation (1) into \(f(\alpha) = 0\):**
- Rearranging equation (1):
\[ \alpha^2 = -2\alpha - 9 \]
- Substitute \(\alpha^2\) in \(f(\alpha)\):
\[ f(\alpha) = a(-2\alpha - 9) + b\alpha + c = 0 \]
- Expanding this gives:
\[ -2a\alpha - 9a + b\alpha + c = 0 \]
- Rearranging terms:
\[ (-2a + b)\alpha + (c - 9a) = 0 \]

5. **Set coefficients to zero:**
- Since this equation must hold for all values of \(\alpha\), both coefficients must equal zero:
\[ -2a + b = 0 \quad \text{(2)} \]
\[ c - 9a = 0 \quad \text{(3)} \]

6. **Solve the system of equations:**
- From equation (2):
\[ b = 2a \]
- From equation (3):
\[ c = 9a \]

7. **Express the ratios:**
- Now we can express the ratios of \(a\), \(b\), and \(c\):
\[ a : b : c = a : 2a : 9a = 1 : 2 : 9 \]

### Final Answer:
The ratio of \(a\), \(b\), and \(c\) is:
\[ \boxed{1 : 2 : 9} \]