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 know that one root is greater than 2 and the other root is less than 2. This means that the quadratic must cross the x-axis at two points, one on either side of x = 2.

### Step 1: Evaluate the function at x = 2

We will substitute \( x = 2 \) into the quadratic equation to find the condition for the roots.

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

Calculating \( f(2) \):

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

### Step 2: Simplify the expression

Now, simplify the expression:

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

### Step 3: Set the condition for the roots

Since one root is greater than 2 and the other is less than 2, we need:

\[ f(2) < 0 \]

This gives us the inequality:

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

### Step 4: Factor the quadratic inequality

Next, we will factor the quadratic expression \( k^2 - k - 6 \):

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

### Step 5: Solve the inequality

Now we will solve the inequality:

\[ (k - 3)(k + 2) < 0 \]

To find the intervals where this inequality holds, we can find the roots of the equation \( (k - 3)(k + 2) = 0 \). The roots are \( k = 3 \) and \( k = -2 \).

### Step 6: Test intervals

We will test the intervals defined by these roots:

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

### Conclusion

The inequality \( (k - 3)(k + 2) < 0 \) holds true for:

\[ -2 < k < 3 \]

Thus, the value of \( k \) lies between -2 and 3.

### Final Answer

The value of \( k \) lies between \( -2 \) and \( 3 \).

---