For what values of `x in R`, the following expressions are positive
i) `x^(2)- 5x + 6`

Answer

To determine the values of \( x \) in \( \mathbb{R} \) for which the expression \( x^2 - 5x + 6 \) is positive, we can follow these steps:

### Step 1: Set the expression greater than zero
We start by setting the expression greater than zero:
\[ x^2 - 5x + 6 > 0 \]

### Step 2: Factor the quadratic expression
Next, we need to factor the quadratic expression. We look for two numbers that multiply to \( 6 \) (the constant term) and add to \( -5 \) (the coefficient of \( x \)). The numbers \( -2 \) and \( -3 \) fit this requirement. Thus, we can factor the expression as:
\[ (x - 2)(x - 3) > 0 \]

### Step 3: Identify the critical points
The critical points occur where the expression equals zero:
\[ (x - 2)(x - 3) = 0 \]
This gives us the points \( x = 2 \) and \( x = 3 \).

### Step 4: Test intervals around the critical points
We will test the sign of the expression in the intervals determined by the critical points:
1. \( (-\infty, 2) \)
2. \( (2, 3) \)
3. \( (3, +\infty) \)

- **Interval \( (-\infty, 2) \)**: Choose \( x = 0 \):
\[ (0 - 2)(0 - 3) = (-2)(-3) = 6 > 0 \]
- **Interval \( (2, 3) \)**: Choose \( x = 2.5 \):
\[ (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25 < 0 \]
- **Interval \( (3, +\infty) \)**: Choose \( x = 4 \):
\[ (4 - 2)(4 - 3) = (2)(1) = 2 > 0 \]

### Step 5: Combine the results
From our tests, we find:
- The expression is positive in the intervals \( (-\infty, 2) \) and \( (3, +\infty) \).
- At the points \( x = 2 \) and \( x = 3 \), the expression equals zero.

### Conclusion
Thus, the values of \( x \) for which \( x^2 - 5x + 6 > 0 \) are:
\[ x \in (-\infty, 2) \cup (3, +\infty) \]