`ax^2 + bx + c = 0(a > 0),` has two roots `alpha and beta` such `alpha < -2 and beta > 2,` then

Options

A. `b^(2)-4acgt0`
B. `clt0`
C. `a+|b|+clt0`
D. `4a+2|b|+clt0`

Explanation

To solve the problem step by step, we will analyze the quadratic equation \( ax^2 + bx + c = 0 \) under the given conditions.

### Step 1: Understand the conditions
We know that:
- The roots of the equation are \( \alpha \) and \( \beta \).
- \( \alpha < -2 \) and \( \beta > 2 \).
- \( a > 0 \).

### Step 2: Analyze the function at specific points
We will evaluate the function \( f(x) = ax^2 + bx + c \) at specific points to derive conditions on \( b \) and \( c \).

### Step 3: Evaluate \( f(0) \)
Calculating \( f(0) \):
\[ f(0) = c \]
Since \( f(0) < 0 \) (because the parabola opens upwards and has roots on either side of the y-axis), we have:
\[ c < 0 \]

### Step 4: Evaluate \( f(2) \)
Next, we evaluate \( f(2) \):
\[ f(2) = 4a + 2b + c \]
Given that \( \beta > 2 \), it implies that the function must be negative at \( x = 2 \):
\[ f(2) < 0 \implies 4a + 2b + c < 0 \]

### Step 5: Evaluate \( f(-2) \)
Now, we evaluate \( f(-2) \):
\[ f(-2) = 4a - 2b + c \]
Since \( \alpha < -2 \), it implies that the function must also be negative at \( x = -2 \):
\[ f(-2) < 0 \implies 4a - 2b + c < 0 \]

### Step 6: Derive conditions from inequalities
We now have two inequalities:
1. \( 4a + 2b + c < 0 \)
2. \( 4a - 2b + c < 0 \)

### Step 7: Subtract the inequalities
Subtract the second inequality from the first:
\[ (4a + 2b + c) - (4a - 2b + c) < 0 \]
This simplifies to:
\[ 4b < 0 \implies b < 0 \]

### Step 8: Use the conditions on \( c \)
Now, since \( c < 0 \) and we have \( b < 0 \), we can summarize the conditions:
- \( c < 0 \)
- \( b < 0 \)

### Step 9: Evaluate the discriminant
For the quadratic to have real roots, the discriminant must be positive:
\[ D = b^2 - 4ac > 0 \]

### Conclusion
Thus, the conditions we derived are:
1. \( b < 0 \)
2. \( c < 0 \)
3. \( b^2 - 4ac > 0 \)