The value of k for which
`2x^(2) + 5xy + 3y^(2) + 3x + 4y + k = 0`
represent pair of straight lines is

Options

A. 0
B. `-1`
C. 1
D. None of these

Answer

To find the value of \( k \) for which the equation

\[ 2x^2 + 5xy + 3y^2 + 3x + 4y + k = 0 \]

represents a pair of straight lines, we will use the condition for the representation of a pair of straight lines in the general form of the conic section.

### Step 1: Identify the coefficients
The general form of the conic section is given by:

\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]

From the given equation, we can identify the coefficients:
- \( a = 2 \) (coefficient of \( x^2 \))
- \( b = 3 \) (coefficient of \( y^2 \))
- \( c = k \) (constant term)
- \( 2h = 5 \) which gives \( h = \frac{5}{2} \) (coefficient of \( xy \))
- \( 2g = 3 \) which gives \( g = \frac{3}{2} \) (coefficient of \( x \))
- \( 2f = 4 \) which gives \( f = 2 \) (coefficient of \( y \))

### Step 2: Apply the condition for pair of straight lines
The condition for the equation to represent a pair of straight lines is given by:

\[ \Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \]

Substituting the values we identified:

\[ \Delta = (2)(3)(k) + 2(2)(\frac{3}{2})(\frac{5}{2}) - (2)(2^2) - (3)(\frac{3}{2})^2 - (k)(\frac{5}{2})^2 = 0 \]

### Step 3: Simplify the expression
Calculating each term:

1. \( abc = 6k \)
2. \( 2fgh = 2 \times 2 \times \frac{3}{2} \times \frac{5}{2} = 30 \)
3. \( af^2 = 2 \times 4 = 8 \)
4. \( bg^2 = 3 \times \frac{9}{4} = \frac{27}{4} \)
5. \( ch^2 = k \times \frac{25}{4} = \frac{25k}{4} \)

Putting it all together:

\[ 6k + 30 - 8 - \frac{27}{4} - \frac{25k}{4} = 0 \]

### Step 4: Combine like terms
To combine the terms, we will express everything with a common denominator of 4:

\[ \frac{24k}{4} + 30 - 8 - \frac{27}{4} - \frac{25k}{4} = 0 \]

This simplifies to:

\[ \frac{(24k - 25k)}{4} + 30 - 8 - \frac{27}{4} = 0 \]

\[ \frac{-k}{4} + 22 - \frac{27}{4} = 0 \]

### Step 5: Solve for \( k \)
Now, let's isolate \( k \):

\[ -k + 88 - 27 = 0 \quad \text{(multiplying through by 4)} \]

\[ -k + 61 = 0 \]

\[ k = 61 \]

Thus, the value of \( k \) for which the equation represents a pair of straight lines is:

\[ \boxed{61} \]