If one root is greater than 2 and the other root is less than 2 for the equation `x^(2)-(k+1)x+(k^(2)+k-8)=0`, then the value of k lies between

Options

A. `(-2, 2)`
B. `(-2, 4)`
C. `(-2, 0)`
D. `(-2, 3)`

Answer

To solve the problem, we need to analyze the quadratic equation given by:

\[ x^2 - (k+1)x + (k^2 + k - 8) = 0 \]

We are told that one root is greater than 2 and the other root is less than 2. This implies that the quadratic must take a negative value at \( x = 2 \).

### Step 1: Evaluate the quadratic at \( x = 2 \)

Substituting \( x = 2 \) into the equation:

\[ f(2) = 2^2 - (k + 1) \cdot 2 + (k^2 + k - 8) \]

Calculating this gives:

\[ f(2) = 4 - 2(k + 1) + (k^2 + k - 8) \]

### Step 2: Simplify the expression

Now, we simplify the expression:

\[ f(2) = 4 - 2k - 2 + k^2 + k - 8 \]
\[ = k^2 - k - 6 \]

### Step 3: Set the inequality

Since we need \( f(2) < 0 \):

\[ k^2 - k - 6 < 0 \]

### Step 4: Factor the quadratic

Next, we factor the quadratic:

\[ k^2 - k - 6 = (k - 3)(k + 2) \]

### Step 5: Determine the intervals

Now we need to find the intervals where this product is negative. The roots of the equation are \( k = -2 \) and \( k = 3 \).

### Step 6: Test intervals

We will test the intervals determined by the roots:

1. For \( k < -2 \): Choose \( k = -3 \):
\[ (-3 - 3)(-3 + 2) = (-6)(-1) = 6 > 0 \]

2. For \( -2 < k < 3 \): Choose \( k = 0 \):
\[ (0 - 3)(0 + 2) = (-3)(2) = -6 < 0 \]

3. For \( k > 3 \): Choose \( k = 4 \):
\[ (4 - 3)(4 + 2) = (1)(6) = 6 > 0 \]

### Conclusion

The quadratic \( (k - 3)(k + 2) < 0 \) holds true in the interval:

\[ -2 < k < 3 \]

Thus, the value of \( k \) lies between \( -2 \) and \( 3 \), not including the endpoints.

### Final Answer

The value of \( k \) lies in the interval \( (-2, 3) \).