Consider the inequation `x^(2) + x + a - 9 lt 0`
The value of the parameter a so that the given inequaiton is ture `AA x in (-1,3)`

Options

A. `(-oo,-3]`
B. `(-3,oo)`
C. `[9,oo)`
D. `(-oo, 34//4)`

Explanation

To solve the inequation \( x^2 + x + a - 9 < 0 \) for the parameter \( a \) such that the inequation holds true for all \( x \) in the interval \( (-1, 3) \), we will follow these steps:

### Step 1: Define the function
Let \( f(x) = x^2 + x + a - 9 \). We need to find the values of \( a \) such that \( f(x) < 0 \) for all \( x \) in the interval \( (-1, 3) \).

### Step 2: Evaluate \( f(x) \) at the endpoints of the interval
We will evaluate \( f(x) \) at \( x = -1 \) and \( x = 3 \).

1. **At \( x = -1 \)**:
\[ f(-1) = (-1)^2 + (-1) + a - 9 = 1 - 1 + a - 9 = a - 9 \]
For \( f(-1) < 0 \):
\[ a - 9 < 0 \implies a < 9 \]

2. **At \( x = 3 \)**:
\[ f(3) = 3^2 + 3 + a - 9 = 9 + 3 + a - 9 = a + 3 \]
For \( f(3) < 0 \):
\[ a + 3 < 0 \implies a < -3 \]

### Step 3: Combine the conditions
We have two conditions:
1. \( a < 9 \)
2. \( a < -3 \)

The more restrictive condition is \( a < -3 \).

### Step 4: Conclusion
Thus, the values of the parameter \( a \) such that the inequation \( x^2 + x + a - 9 < 0 \) holds for all \( x \) in the interval \( (-1, 3) \) is:
\[ a < -3 \]

### Final Answer
The value of the parameter \( a \) is \( (-\infty, -3) \).

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