Find the value of k so that the sum of the roots of the quadratic equation is equal to the product to the roots :
`(k+1)x^(2)+2kx+4=0`

Options

A. `-2`
B. `-4`
C. 6
D. 8

Answer

To solve the problem, we need to find the value of \( k \) such that the sum of the roots of the quadratic equation \( (k+1)x^2 + 2kx + 4 = 0 \) is equal to the product of the roots.

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given quadratic equation is in the form \( ax^2 + bx + c = 0 \), where:
- \( a = k + 1 \)
- \( b = 2k \)
- \( c = 4 \)

2. **Sum of the roots**:
The sum of the roots \( \alpha + \beta \) of a quadratic equation is given by the formula:
\[ \alpha + \beta = -\frac{b}{a} \]
Substituting the values of \( b \) and \( a \):
\[ \alpha + \beta = -\frac{2k}{k + 1} \]

3. **Product of the roots**:
The product of the roots \( \alpha \beta \) is given by:
\[ \alpha \beta = \frac{c}{a} \]
Substituting the values of \( c \) and \( a \):
\[ \alpha \beta = \frac{4}{k + 1} \]

4. **Set the sum equal to the product**:
According to the problem, we need to set the sum of the roots equal to the product of the roots:
\[ -\frac{2k}{k + 1} = \frac{4}{k + 1} \]

5. **Cross-multiply**:
To eliminate the fractions, we cross-multiply:
\[ -2k(k + 1) = 4(k + 1) \]

6. **Expand both sides**:
Expanding both sides gives:
\[ -2k^2 - 2k = 4k + 4 \]

7. **Rearrange the equation**:
Bringing all terms to one side results in:
\[ -2k^2 - 2k - 4k - 4 = 0 \]
Simplifying this, we have:
\[ -2k^2 - 6k - 4 = 0 \]

8. **Divide by -2**:
To make the equation simpler, divide everything by -2:
\[ k^2 + 3k + 2 = 0 \]

9. **Factor the quadratic**:
We can factor the quadratic equation:
\[ (k + 1)(k + 2) = 0 \]

10. **Find the roots**:
Setting each factor to zero gives:
\[ k + 1 = 0 \quad \Rightarrow \quad k = -1 \]
\[ k + 2 = 0 \quad \Rightarrow \quad k = -2 \]

### Final Answer:
The values of \( k \) are \( k = -1 \) and \( k = -2 \).