Solve `x^(2)-7x+6 gt 0`.

Answer

To solve the inequality \( x^2 - 7x + 6 > 0 \), we will follow these steps:

### Step 1: Factor the quadratic expression
We start by factoring the quadratic expression \( x^2 - 7x + 6 \).

We need to find two numbers that multiply to \( 6 \) (the constant term) and add up to \( -7 \) (the coefficient of \( x \)). The numbers that satisfy this are \( -6 \) and \( -1 \).

Thus, we can factor the expression as:
\[ x^2 - 7x + 6 = (x - 6)(x - 1) \]

### Step 2: Set the factors greater than zero
Now we rewrite the inequality:
\[ (x - 6)(x - 1) > 0 \]

### Step 3: Identify the critical points
The critical points occur where each factor equals zero:
1. \( x - 6 = 0 \) → \( x = 6 \)
2. \( x - 1 = 0 \) → \( x = 1 \)

### Step 4: Test intervals around the critical points
We will test the sign of the product \( (x - 6)(x - 1) \) in the intervals defined by the critical points \( x = 1 \) and \( x = 6 \):
- Interval 1: \( (-\infty, 1) \)
- Interval 2: \( (1, 6) \)
- Interval 3: \( (6, \infty) \)

**Testing Interval 1: \( x < 1 \) (e.g., \( x = 0 \))**
\[ (0 - 6)(0 - 1) = (-6)(-1) = 6 > 0 \quad \text{(satisfied)} \]

**Testing Interval 2: \( 1 < x < 6 \) (e.g., \( x = 2 \))**
\[ (2 - 6)(2 - 1) = (-4)(1) = -4 < 0 \quad \text{(not satisfied)} \]

**Testing Interval 3: \( x > 6 \) (e.g., \( x = 7 \))**
\[ (7 - 6)(7 - 1) = (1)(6) = 6 > 0 \quad \text{(satisfied)} \]

### Step 5: Combine the results
The inequality \( (x - 6)(x - 1) > 0 \) is satisfied in the intervals:
1. \( (-\infty, 1) \)
2. \( (6, \infty) \)

### Final Solution
The solution to the inequality \( x^2 - 7x + 6 > 0 \) is:
\[ x < 1 \quad \text{or} \quad x > 6 \]