If x is real and `x^(2) - 3x + 2 gt 0, x^(2)- 3x - 4 le 0,` then which one of the following is correct?

Options

A. ` -1 le x le 4 `
B. ` 2 le x le 4 `
C. ` -1 lt x lt 1 `
D. ` -1 le x lt 1 or 2 lt x le 4 `

Explanation

To solve the inequalities \( x^2 - 3x + 2 > 0 \) and \( x^2 - 3x - 4 \leq 0 \), we will break down the solution step by step.

### 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 sign of the expression in the intervals defined by the critical points \( (-\infty, 1) \), \( (1, 2) \), and \( (2, \infty) \):
- For \( x < 1 \) (e.g., \( x = 0 \)):
\[ (0 - 1)(0 - 2) = 1 \quad (\text{positive}) \]
- For \( 1 < x < 2 \) (e.g., \( x = 1.5 \)):
\[ (1.5 - 1)(1.5 - 2) = -0.25 \quad (\text{negative}) \]
- For \( x > 2 \) (e.g., \( x = 3 \)):
\[ (3 - 1)(3 - 2) = 2 \quad (\text{positive}) \]
4. The solution for \( x^2 - 3x + 2 > 0 \) 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 sign of the expression in the intervals defined by the critical points \( (-\infty, -1) \), \( (-1, 4) \), and \( (4, \infty) \):
- For \( x < -1 \) (e.g., \( x = -2 \)):
\[ (-2 - 4)(-2 + 1) = 6 \quad (\text{positive}) \]
- For \( -1 < x < 4 \) (e.g., \( x = 0 \)):
\[ (0 - 4)(0 + 1) = -4 \quad (\text{negative}) \]
- For \( x > 4 \) (e.g., \( x = 5 \)):
\[ (5 - 4)(5 + 1) = 6 \quad (\text{positive}) \]
4. The solution for \( x^2 - 3x - 4 \leq 0 \) 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. The common intervals are:
- From the first interval \( (-\infty, 1) \) and the second interval \( [-1, 4] \):
\[ x \in [-1, 1) \]
- From the second interval \( [2, 4] \):
\[ x \in (2, 4] \]

### Final Solution

Combining these intervals, we have:
\[ x \in [-1, 1) \cup (2, 4] \]