If `0 lt a lt b lt c` and equation `ax^(2)+bx+c=0` does not posses distinct roots (say `alpha` and `beta`), then

Options

A. `|alpha|=|beta|`
B. `|alpha|gt1`
C. `|beta|lt 1`
D. none of these

Answer

To solve the problem, we start by analyzing the given quadratic equation and the conditions provided.

### Step 1: Understanding the Conditions
We have the quadratic equation:
\[ ax^2 + bx + c = 0 \]
with the conditions:
\[ 0 < a < b < c \]
This means that \( a, b, c \) are all positive and \( c \) is the largest, followed by \( b \), and \( a \) is the smallest.

### Step 2: Roots of the Quadratic Equation
The roots of the quadratic equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

### Step 3: Condition for Non-distinct Roots
The equation does not possess distinct roots, which means the discriminant must be zero:
\[ b^2 - 4ac = 0 \]
This implies that the roots are equal, denoted as \( \alpha = \beta \).

### Step 4: Finding the Roots
Since the roots are equal, we can express them as:
\[ \alpha = \beta = \frac{-b}{2a} \]
This shows that both roots are the same.

### Step 5: Analyzing the Roots
Given that \( a, b, c > 0 \) and \( b > a \), we can analyze the value of \( \alpha \):
- Since \( b \) is positive and \( a \) is positive, the expression \( -b/(2a) \) will be negative.
- Therefore, both roots \( \alpha \) and \( \beta \) are negative.

### Step 6: Modulus of the Roots
Since both roots are negative, we can find their modulus:
\[ |\alpha| = |\beta| = -\frac{-b}{2a} = \frac{b}{2a} \]
Since \( b > a \), it follows that:
\[ \frac{b}{2a} > 1 \]
Thus, we conclude:
\[ |\alpha| > 1 \quad \text{and} \quad |\beta| > 1 \]

### Conclusion
Based on the analysis, we can summarize that:
- The roots \( \alpha \) and \( \beta \) are equal and negative.
- Their moduli are greater than 1.