If `x^(2)+6x-27 gt 0 and x^(2)-3x-4 lt 0` , then :

Options

A. ` x gt 3`
B. `- 1 lt x lt 4`
C. `3 lt x lt 4 `
D. `x lt -9 or x gt -1`

Answer

To solve the inequalities \( x^2 + 6x - 27 > 0 \) and \( x^2 - 3x - 4 < 0 \), we will follow these steps:

### Step 1: Solve the first inequality \( x^2 + 6x - 27 > 0 \)

1. **Factor the quadratic expression**:
\[ x^2 + 6x - 27 = (x - 3)(x + 9) \]
We can find the roots by setting the equation to zero:
\[ x^2 + 6x - 27 = 0 \implies (x - 3)(x + 9) = 0 \]
The roots are \( x = 3 \) and \( x = -9 \).

2. **Determine the intervals**:
The critical points divide the number line into intervals:
- \( (-\infty, -9) \)
- \( (-9, 3) \)
- \( (3, \infty) \)

3. **Test each interval**:
- For \( x < -9 \) (e.g., \( x = -10 \)):
\[ (-10 - 3)(-10 + 9) = (-13)(-1) > 0 \quad \text{(True)} \]
- For \( -9 < x < 3 \) (e.g., \( x = 0 \)):
\[ (0 - 3)(0 + 9) = (-3)(9) < 0 \quad \text{(False)} \]
- For \( x > 3 \) (e.g., \( x = 4 \)):
\[ (4 - 3)(4 + 9) = (1)(13) > 0 \quad \text{(True)} \]

4. **Conclusion for the first inequality**:
The solution is:
\[ x \in (-\infty, -9) \cup (3, \infty) \]

### Step 2: Solve the second inequality \( x^2 - 3x - 4 < 0 \)

1. **Factor the quadratic expression**:
\[ x^2 - 3x - 4 = (x - 4)(x + 1) \]
Setting the equation to zero gives:
\[ x^2 - 3x - 4 = 0 \implies (x - 4)(x + 1) = 0 \]
The roots are \( x = 4 \) and \( x = -1 \).

2. **Determine the intervals**:
The critical points divide the number line into intervals:
- \( (-\infty, -1) \)
- \( (-1, 4) \)
- \( (4, \infty) \)

3. **Test each interval**:
- For \( x < -1 \) (e.g., \( x = -2 \)):
\[ (-2 - 4)(-2 + 1) = (-6)(-1) > 0 \quad \text{(False)} \]
- For \( -1 < x < 4 \) (e.g., \( x = 0 \)):
\[ (0 - 4)(0 + 1) = (-4)(1) < 0 \quad \text{(True)} \]
- For \( x > 4 \) (e.g., \( x = 5 \)):
\[ (5 - 4)(5 + 1) = (1)(6) > 0 \quad \text{(False)} \]

4. **Conclusion for the second inequality**:
The solution is:
\[ x \in (-1, 4) \]

### Step 3: Find the intersection of the two solutions

- From the first inequality, we have:
\[ x \in (-\infty, -9) \cup (3, \infty) \]
- From the second inequality, we have:
\[ x \in (-1, 4) \]

The intersection of these two sets is:
- The interval \( (3, 4) \).

### Final Solution
Thus, the solution to the given inequalities is:
\[ x \in (3, 4) \]