To solve the inequalities \(x^2 + 5x + 6 \geq 0\) and \(x^2 + 3x - 4 < 0\), we will proceed step by step.
### Step 1: Solve the first inequality \(x^2 + 5x + 6 \geq 0\)
1. **Factor the quadratic expression**:
\[
x^2 + 5x + 6 = (x + 2)(x + 3)
\]
Thus, we rewrite the inequality:
\[
(x + 2)(x + 3) \geq 0
\]
2. **Determine the critical points**:
The critical points are found by setting each factor to zero:
\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\]
\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]
3. **Test intervals**:
We will test the intervals determined by the critical points: \((- \infty, -3)\), \((-3, -2)\), and \((-2, \infty)\).
- For \(x < -3\) (e.g., \(x = -4\)):
\((x + 2)(x + 3) = (-4 + 2)(-4 + 3) = (-2)(-1) = 2 \geq 0\) (True)
- For \(-3 < x < -2\) (e.g., \(x = -2.5\)):
\((x + 2)(x + 3) = (-2.5 + 2)(-2.5 + 3) = (-0.5)(0.5) = -0.25 < 0\) (False)
- For \(x > -2\) (e.g., \(x = 0\)):
\((x + 2)(x + 3) = (0 + 2)(0 + 3) = (2)(3) = 6 \geq 0\) (True)
4. **Combine results**:
The solution for the first inequality is:
\[
x \in (-\infty, -3] \cup [-2, \infty)
\]
### Step 2: Solve the second inequality \(x^2 + 3x - 4 < 0\)
1. **Factor the quadratic expression**:
\[
x^2 + 3x - 4 = (x + 4)(x - 1)
\]
Thus, we rewrite the inequality:
\[
(x + 4)(x - 1) < 0
\]
2. **Determine the critical points**:
The critical points are:
\[
x + 4 = 0 \quad \Rightarrow \quad x = -4
\]
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]
3. **Test intervals**:
We will test the intervals determined by the critical points: \((- \infty, -4)\), \((-4, 1)\), and \((1, \infty)\).
- For \(x < -4\) (e.g., \(x = -5\)):
\((x + 4)(x - 1) = (-5 + 4)(-5 - 1) = (-1)(-6) = 6 > 0\) (False)
- For \(-4 < x < 1\) (e.g., \(x = 0\)):
\((x + 4)(x - 1) = (0 + 4)(0 - 1) = (4)(-1) = -4 < 0\) (True)
- For \(x > 1\) (e.g., \(x = 2\)):
\((x + 4)(x - 1) = (2 + 4)(2 - 1) = (6)(1) = 6 > 0\) (False)
4. **Combine results**:
The solution for the second inequality is:
\[
x \in (-4, 1)
\]
### Step 3: Find the intersection of the two solutions
We need to find the values of \(x\) that satisfy both inequalities:
1. From the first inequality: \(x \in (-\infty, -3] \cup [-2, \infty)\)
2. From the second inequality: \(x \in (-4, 1)\)
**Intersection**:
- The interval \((-4, -3]\) is common in both solutions.
- The interval \([-2, \infty)\) does not overlap with \((-4, 1)\).
Thus, the final solution set is:
\[
x \in (-4, -3]
\]
