Let a,b,c be real if `ax^(2)+bx+c=0` has two real roots `alpha` and `beta` where `a lt -2` and `beta gt2`, then

Options

A. `4+(2b)/a+c/a=0`
B. `4-(2b)/a+c/a=0`
C. `4+(2b)/a-c/alt0`
D. `4-(2b)/a+c/alt0`

Answer

To solve the problem, we need to analyze the quadratic equation \( ax^2 + bx + c = 0 \) given the conditions on its roots \( \alpha \) and \( \beta \).

### Step 1: Understand the conditions on the roots
We know that:
- \( \alpha < -2 \)
- \( \beta > 2 \)

This means that the roots of the quadratic equation are positioned such that one root is less than -2 and the other root is greater than 2.

### Step 2: Analyze the quadratic function
The quadratic function can be expressed as:
\[ f(x) = ax^2 + bx + c \]
Since \( a < -2 \), the parabola opens downwards (as \( a \) is negative).

### Step 3: Determine the behavior of the function between the roots
Since \( \alpha < -2 \) and \( \beta > 2 \), the function will be negative between the roots. Therefore, for any \( x \) in the interval \( (-2, 2) \), we have:
\[ f(x) < 0 \]

### Step 4: Evaluate the function at the endpoints
We need to evaluate \( f(-2) \) and \( f(2) \):
1. **At \( x = 2 \)**:
\[ f(2) = a(2^2) + b(2) + c = 4a + 2b + c \]
Since \( f(x) < 0 \) for \( x \) between the roots, we have:
\[ 4a + 2b + c < 0 \]

2. **At \( x = -2 \)**:
\[ f(-2) = a(-2^2) + b(-2) + c = 4a - 2b + c \]
Similarly, since \( f(x) < 0 \) for \( x \) between the roots, we have:
\[ 4a - 2b + c < 0 \]

### Step 5: Analyze the implications of the inequalities
We now have two inequalities:
1. \( 4a + 2b + c < 0 \)
2. \( 4a - 2b + c < 0 \)

### Step 6: Solve the inequalities
Subtract the second inequality from the first:
\[ (4a + 2b + c) - (4a - 2b + c) < 0 \]
This simplifies to:
\[ 4b < 0 \implies b < 0 \]

### Conclusion
From our analysis, we conclude that:
- \( a < -2 \)
- \( b < 0 \)

Thus, the conditions on the coefficients \( a \) and \( b \) are established based on the given conditions of the roots.