If a, b, c are real if `ax^(2)+ bx + c = 0` has two real roots `alpha, beta` where `a lt -1, beta gt 1` then

Options

A. `1+ (c) /(a)+ |(b)/(a)| lt 0`
B. `1- (c) /(a)+ |(b)/(a)| lt 0`
C. c=0
D. `(c) /(a)+ |(b)/(a)| -1 gt 0`

Answer

To solve the problem, we need to analyze the given quadratic equation \( ax^2 + bx + c = 0 \) under the conditions provided. Let's break down the solution step by step.

### Step 1: Understanding the Quadratic Function
The quadratic function is given as:
\[ f(x) = ax^2 + bx + c \]
where \( a, b, c \) are real numbers. The roots of this function are denoted as \( \alpha \) and \( \beta \).

### Step 2: Conditions on the Roots
We know from the problem statement that:
- \( \alpha < -1 \)
- \( \beta > 1 \)

### Step 3: Nature of the Parabola
Since \( a < -1 \), the parabola opens downwards (because \( a < 0 \)). This means that the vertex of the parabola is a maximum point.

### Step 4: Evaluating the Function at Specific Points
We will evaluate the function at \( x = -1 \) and \( x = 1 \):
1. **At \( x = -1 \)**:
\[ f(-1) = a(-1)^2 + b(-1) + c = a - b + c \]
2. **At \( x = 1 \)**:
\[ f(1) = a(1)^2 + b(1) + c = a + b + c \]

### Step 5: Conditions on the Function Values
Since \( \alpha < -1 \) and \( \beta > 1 \), it implies:
- The function must be positive between the roots, meaning \( f(-1) < 0 \) and \( f(1) < 0 \).

Thus, we have:
1. \( a - b + c < 0 \)
2. \( a + b + c < 0 \)

### Step 6: Combining the Inequalities
From the inequalities:
1. \( a - b + c < 0 \)
2. \( a + b + c < 0 \)

We can manipulate these inequalities to find conditions on \( b \) and \( c \):
- From the first inequality, rearranging gives:
\[ c < b - a \]
- From the second inequality, rearranging gives:
\[ c < -a - b \]

### Step 7: Conclusion
The conditions derived suggest that \( c \) must be less than both \( b - a \) and \( -a - b \). This indicates that the value of \( c \) is constrained by the values of \( a \) and \( b \).

### Final Answer
The correct option based on the analysis is:
- \( 1 + \frac{c}{a} + \left| \frac{b}{a} \right| < 0 \)