The expression `x^2 + 2xy + ky^2 + 2x + k = 0` can be resolved into two linear factors, then `k in`

Answer

To solve the problem, we need to determine the values of \( k \) for which the expression \( x^2 + 2xy + ky^2 + 2x + k = 0 \) can be factored into two linear factors. This condition implies that the expression represents a pair of straight lines, which can be analyzed using the discriminant condition.

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given expression is \( x^2 + 2xy + ky^2 + 2x + k = 0 \). We can identify the coefficients as follows:
- \( a = 1 \) (coefficient of \( x^2 \))
- \( b = k \) (coefficient of \( y^2 \))
- \( c = k \) (constant term)
- \( f = 1 \) (coefficient of \( xy \))
- \( g = 2 \) (coefficient of \( x \))
- \( h = 0 \) (coefficient of \( y \))

2. **Use the condition for a pair of straight lines**:
The condition for the expression to represent a pair of straight lines is given by the determinant \( \Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \).

3. **Substitute the coefficients into the determinant**:
Substituting the identified coefficients into the determinant formula:
\[ \Delta = (1)(k)(k) + 2(1)(2)(0) - (1)(1^2) - (k)(2^2) - (k)(0^2) \]
Simplifying this gives:
\[ \Delta = k^2 - 1 - 4k = k^2 - 4k - 1 \]

4. **Set the determinant equal to zero**:
To find the values of \( k \), we set the determinant equal to zero:
\[ k^2 - 4k - 1 = 0 \]

5. **Solve the quadratic equation**:
We can solve this quadratic equation using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -4 \), and \( c = -1 \).
\[ k = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-1)}}{2(1)} = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm \sqrt{20}}{2} \]
Simplifying \( \sqrt{20} \):
\[ \sqrt{20} = 2\sqrt{5} \]
Thus,
\[ k = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5} \]

6. **Final values of \( k \)**:
Therefore, the values of \( k \) for which the expression can be factored into linear factors are:
\[ k = 2 + \sqrt{5} \quad \text{or} \quad k = 2 - \sqrt{5} \]