If the product of two of the roots of `x^3 +kx ^2 -3x+4=0` is -1 then k=

Options

A. `7/2 `
B. `9/2`
C. `-7/2`
D. `-9/2`

Answer

To solve the problem, we need to find the value of \( k \) in the polynomial equation \( x^3 + kx^2 - 3x + 4 = 0 \) given that the product of two of its roots is \( -1 \).

### Step-by-step Solution:

1. **Identify the roots**: Let the roots of the polynomial be \( \alpha, \beta, \) and \( \gamma \).

2. **Use Vieta's Formulas**: According to Vieta's formulas for a cubic equation \( ax^3 + bx^2 + cx + d = 0 \):
- The sum of the roots \( \alpha + \beta + \gamma = -\frac{b}{a} = -k \).
- The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = -3 \).
- The product of the roots \( \alpha\beta\gamma = -\frac{d}{a} = -4 \).

3. **Given Condition**: We know that \( \alpha \beta = -1 \).

4. **Find \( \gamma \)**: Using the product of the roots:
\[ \alpha \beta \gamma = -4 \]
Substituting \( \alpha \beta = -1 \):
\[ -1 \cdot \gamma = -4 \implies \gamma = 4 \]

5. **Calculate \( \alpha + \beta \)**: From the sum of the roots:
\[ \alpha + \beta + \gamma = -k \]
Substituting \( \gamma = 4 \):
\[ \alpha + \beta + 4 = -k \implies \alpha + \beta = -k - 4 \]

6. **Use the sum of products**: From the sum of the products of the roots:
\[ \alpha \beta + \beta \gamma + \gamma \alpha = -3 \]
Substituting \( \alpha \beta = -1 \) and \( \gamma = 4 \):
\[ -1 + 4\beta + 4\alpha = -3 \]
Rearranging gives:
\[ 4(\alpha + \beta) - 1 = -3 \implies 4(\alpha + \beta) = -2 \implies \alpha + \beta = -\frac{1}{2} \]

7. **Substitute back to find \( k \)**: Now substituting \( \alpha + \beta = -\frac{1}{2} \) into the equation \( \alpha + \beta = -k - 4 \):
\[ -\frac{1}{2} = -k - 4 \]
Rearranging gives:
\[ -k = -\frac{1}{2} + 4 \implies -k = -\frac{1}{2} + \frac{8}{2} \implies -k = \frac{7}{2} \]
Therefore, \( k = -\frac{7}{2} \).

### Final Answer:
\[ k = -\frac{7}{2} \]