Find the value of k for which one root of the equation of `x^(2)-(k+1)x+k^(2)+k-8=0` exceed 2 and other is smaller than 2.

Answer

To solve the problem, we need to find the value of \( k \) such that one root of the quadratic equation \( x^2 - (k+1)x + (k^2 + k - 8) = 0 \) is greater than 2 and the other root is less than 2.

### Step-by-Step Solution:

1. **Identify the quadratic equation**:
The given equation is:
\[ x^2 - (k+1)x + (k^2 + k - 8) = 0 \]

2. **Evaluate the function at \( x = 2 \)**:
We need to find \( f(2) \) and ensure that it is less than 0:
\[ f(2) = 2^2 - (k+1) \cdot 2 + (k^2 + k - 8) \]
Simplifying this:
\[ f(2) = 4 - 2(k+1) + (k^2 + k - 8) \]
\[ = 4 - 2k - 2 + k^2 + k - 8 \]
\[ = k^2 - k - 6 \]

3. **Set the inequality**:
We want \( f(2) < 0 \):
\[ k^2 - k - 6 < 0 \]

4. **Factor the quadratic**:
To solve the inequality, we first factor \( k^2 - k - 6 \):
\[ k^2 - k - 6 = (k - 3)(k + 2) \]

5. **Determine the critical points**:
The critical points are \( k = 3 \) and \( k = -2 \).

6. **Test intervals**:
We will test the intervals determined by the critical points:
- For \( k < -2 \) (e.g., \( k = -3 \)):
\[ (-3 - 3)(-3 + 2) = (-6)(-1) > 0 \]
- For \( -2 < k < 3 \) (e.g., \( k = 0 \)):
\[ (0 - 3)(0 + 2) = (-3)(2) < 0 \]
- For \( k > 3 \) (e.g., \( k = 4 \)):
\[ (4 - 3)(4 + 2) = (1)(6) > 0 \]

7. **Conclusion**:
The inequality \( (k - 3)(k + 2) < 0 \) holds true for:
\[ -2 < k < 3 \]
Therefore, the values of \( k \) for which one root is greater than 2 and the other is less than 2 are:
\[ k \in (-2, 3) \]