To solve the inequalities \( x^2 - 3x + 2 > 0 \) and \( x^2 - 3x - 4 \leq 0 \), we will tackle each inequality separately and then find the intersection of the solutions.
### Step 1: Solve the first inequality \( x^2 - 3x + 2 > 0 \)
1. Factor the quadratic expression:
\[
x^2 - 3x + 2 = (x - 1)(x - 2)
\]
2. Set the factors to zero to find the critical points:
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
3. Determine the intervals to test:
- Interval 1: \( (-\infty, 1) \)
- Interval 2: \( (1, 2) \)
- Interval 3: \( (2, \infty) \)
4. Test a point from each interval:
- For \( x = 0 \) in \( (-\infty, 1) \):
\[
(0 - 1)(0 - 2) = 2 > 0 \quad \text{(True)}
\]
- For \( x = 1.5 \) in \( (1, 2) \):
\[
(1.5 - 1)(1.5 - 2) = -0.25 < 0 \quad \text{(False)}
\]
- For \( x = 3 \) in \( (2, \infty) \):
\[
(3 - 1)(3 - 2) = 2 > 0 \quad \text{(True)}
\]
5. The solution for the first inequality is:
\[
x \in (-\infty, 1) \cup (2, \infty)
\]
### Step 2: Solve the second inequality \( x^2 - 3x - 4 \leq 0 \)
1. Factor the quadratic expression:
\[
x^2 - 3x - 4 = (x - 4)(x + 1)
\]
2. Set the factors to zero to find the critical points:
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4
\]
\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\]
3. Determine the intervals to test:
- Interval 1: \( (-\infty, -1) \)
- Interval 2: \( (-1, 4) \)
- Interval 3: \( (4, \infty) \)
4. Test a point from each interval:
- For \( x = -2 \) in \( (-\infty, -1) \):
\[
(-2 - 4)(-2 + 1) = 6 > 0 \quad \text{(False)}
\]
- For \( x = 0 \) in \( (-1, 4) \):
\[
(0 - 4)(0 + 1) = -4 < 0 \quad \text{(True)}
\]
- For \( x = 5 \) in \( (4, \infty) \):
\[
(5 - 4)(5 + 1) = 6 > 0 \quad \text{(False)}
\]
5. The solution for the second inequality is:
\[
x \in [-1, 4]
\]
### Step 3: Find the intersection of the two solutions
1. The first inequality gives us:
\[
x \in (-\infty, 1) \cup (2, \infty)
\]
2. The second inequality gives us:
\[
x \in [-1, 4]
\]
3. Now we find the intersection:
- From \( (-\infty, 1) \) and \( [-1, 4] \):
\[
(-\infty, 1) \cap [-1, 4] = [-1, 1)
\]
- From \( (2, \infty) \) and \( [-1, 4] \):
\[
(2, \infty) \cap [-1, 4] = (2, 4]
\]
4. Therefore, the final solution is:
\[
x \in [-1, 1) \cup (2, 4]
\]
### Conclusion
The correct option is that \( x \) belongs to the intervals \( [-1, 1) \) or \( (2, 4] \).
