If the equation `x^2+2x+3=0` and `ax^2+bx+c=0` have a common root then `a:b:c` is

Options

A. `3:2:1`
B. `1:3:2`
C. `3:1:2`
D. `1:2:3`

Explanation

To solve the problem, we need to find the ratio \( a:b:c \) given that the equations \( x^2 + 2x + 3 = 0 \) and \( ax^2 + bx + c = 0 \) have a common root.

### Step-by-Step Solution:

1. **Identify the first equation**:
The first equation is given as:
\[ x^2 + 2x + 3 = 0 \]

2. **Assume a common root**:
Let \( r \) be the common root of both equations. Since \( r \) is a root of the first equation, we can substitute \( r \) into it:
\[ r^2 + 2r + 3 = 0 \]

3. **Substitute into the second equation**:
Since \( r \) is also a root of the second equation \( ax^2 + bx + c = 0 \), we substitute \( r \) into it:
\[ ar^2 + br + c = 0 \]

4. **Express \( ar^2 + br + c \) using the first equation**:
From the first equation, we know that \( r^2 = -2r - 3 \). We can substitute this expression for \( r^2 \) into the second equation:
\[ a(-2r - 3) + br + c = 0 \]
This simplifies to:
\[ -2ar - 3a + br + c = 0 \]

5. **Rearranging the equation**:
Grouping the terms involving \( r \):
\[ (-2a + b)r + (c - 3a) = 0 \]

6. **Setting coefficients to zero**:
For the equation to hold for all values of \( r \), both coefficients must be zero:
\[ -2a + b = 0 \quad \text{(1)} \]
\[ c - 3a = 0 \quad \text{(2)} \]

7. **Solving the equations**:
From equation (1):
\[ b = 2a \]
From equation (2):
\[ c = 3a \]

8. **Finding the ratio \( a:b:c \)**:
Now we can express \( a, b, c \) in terms of \( a \):
\[ a = a, \quad b = 2a, \quad c = 3a \]
Therefore, the ratio \( a:b:c \) is:
\[ a:b:c = a:2a:3a = 1:2:3 \]

### Final Answer:
Thus, the ratio \( a:b:c \) is:
\[ \boxed{1:2:3} \]