If a and 4a + 3b + 2c have same sign. Then, `ax^(2) + bx + c = 0(a ne 0)` cannot have both roots belonging to

Options

A. `(-1, 2)`
B. `(-1, 1)`
C. (1, 2)
D. `(-2, -1)`

Explanation

To solve the problem, we need to analyze the conditions under which the quadratic equation \( ax^2 + bx + c = 0 \) (where \( a \neq 0 \)) cannot have both roots belonging to the interval \( (1, 2) \) given that \( a \) and \( 4a + 3b + 2c \) have the same sign.

### Step-by-step Solution:

1. **Understanding the Roots**:
Let \( \alpha \) and \( \beta \) be the roots of the quadratic equation. According to Vieta's formulas:
- The sum of the roots: \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots: \( \alpha \beta = \frac{c}{a} \)

2. **Condition for Same Sign**:
Since \( a \) and \( 4a + 3b + 2c \) have the same sign, we can express this mathematically:
\[ a(4a + 3b + 2c) > 0 \]
This means both \( a \) and \( 4a + 3b + 2c \) are either both positive or both negative.

3. **Rearranging the Condition**:
From the inequality \( 4a + 3b + 2c > 0 \), we can express it in terms of \( b \) and \( c \):
\[ 4 + \frac{3b}{a} + \frac{2c}{a} > 0 \]
Let \( \frac{b}{a} = k \) and \( \frac{c}{a} = m \). Then, we rewrite the inequality as:
\[ 4 + 3k + 2m > 0 \]

4. **Expressing in terms of Roots**:
Substitute \( k \) and \( m \) back in terms of roots:
\[ k = -\frac{\alpha + \beta}{a}, \quad m = \frac{\alpha \beta}{a} \]
Thus, the inequality becomes:
\[ 4 - 3\frac{\alpha + \beta}{a} + 2\frac{\alpha \beta}{a} > 0 \]

5. **Analyzing the Roots in the Interval (1, 2)**:
If both roots \( \alpha \) and \( \beta \) belong to the interval \( (1, 2) \), then:
- The sum of the roots \( \alpha + \beta \) would be in the interval \( (2, 4) \).
- The product of the roots \( \alpha \beta \) would be in the interval \( (1, 4) \).

6. **Substituting the Values**:
If \( \alpha + \beta \) is in \( (2, 4) \) and \( \alpha \beta \) is in \( (1, 4) \), we can substitute these values back into our inequality:
- For \( 4 - 3k + 2m > 0 \) to hold, we can analyze the extreme cases of \( k \) and \( m \).

7. **Conclusion**:
If both roots belong to \( (1, 2) \), then the conditions derived from \( 4 + 3k + 2m > 0 \) would not hold true. Therefore, we conclude that if \( a \) and \( 4a + 3b + 2c \) have the same sign, then the quadratic equation \( ax^2 + bx + c = 0 \) cannot have both roots belonging to the interval \( (1, 2) \).