1. **State the problem:** Find the value of $k$ such that the equation $$x^2 + kxy + 2y^2 + 3x + 5y + 2 = 0$$ represents a pair of straight lines. 2. **Formula and rule:** A second-degree equation $$Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$$ represents a pair of straight lines if the determinant $$\Delta = \begin{vmatrix} A & H & G \\ H & B & F \\ G & F & C \end{vmatrix} = 0$$ 3. **Identify coefficients:** Here, $$A=1$$, $$2H = k \Rightarrow H = \frac{k}{2}$$, $$B=2$$, $$2G=3 \Rightarrow G=\frac{3}{2}$$, $$2F=5 \Rightarrow F=\frac{5}{2}$$, $$C=2$$. 4. **Calculate determinant:** $$\Delta = \begin{vmatrix} 1 & \frac{k}{2} & \frac{3}{2} \\ \frac{k}{2} & 2 & \frac{5}{2} \\ \frac{3}{2} & \frac{5}{2} & 2 \end{vmatrix}$$ Expanding: $$\Delta = 1 \times \begin{vmatrix} 2 & \frac{5}{2} \\ \frac{5}{2} & 2 \end{vmatrix} - \frac{k}{2} \times \begin{vmatrix} \frac{k}{2} & \frac{5}{2} \\ \frac{3}{2} & 2 \end{vmatrix} + \frac{3}{2} \times \begin{vmatrix} \frac{k}{2} & 2 \\ \frac{3}{2} & \frac{5}{2} \end{vmatrix}$$ Calculate each minor: - $$M_1 = 2 \times 2 - \frac{5}{2} \times \frac{5}{2} = 4 - \frac{25}{4} = \frac{16}{4} - \frac{25}{4} = -\frac{9}{4}$$ - $$M_2 = \frac{k}{2} \times 2 - \frac{5}{2} \times \frac{3}{2} = k - \frac{15}{4}$$ - $$M_3 = \frac{k}{2} \times \frac{5}{2} - 2 \times \frac{3}{2} = \frac{5k}{4} - 3$$ So, $$\Delta = 1 \times \left(-\frac{9}{4}\right) - \frac{k}{2} \times (k - \frac{15}{4}) + \frac{3}{2} \times \left(\frac{5k}{4} - 3\right)$$ Simplify: $$\Delta = -\frac{9}{4} - \frac{k}{2}k + \frac{k}{2} \times \frac{15}{4} + \frac{3}{2} \times \frac{5k}{4} - \frac{3}{2} \times 3$$ $$= -\frac{9}{4} - \frac{k^2}{2} + \frac{15k}{8} + \frac{15k}{8} - \frac{9}{2}$$ Combine like terms: $$\Delta = -\frac{9}{4} - \frac{9}{2} - \frac{k^2}{2} + \frac{30k}{8}$$ $$= -\frac{9}{4} - \frac{18}{4} - \frac{k^2}{2} + \frac{15k}{4}$$ $$= -\frac{27}{4} - \frac{k^2}{2} + \frac{15k}{4}$$ Multiply entire equation by 4 to clear denominators: $$4\Delta = -27 - 2k^2 + 15k = 0$$ 5. **Solve for $k$:** $$-2k^2 + 15k - 27 = 0$$ Multiply both sides by -1: $$2k^2 - 15k + 27 = 0$$ Use quadratic formula: $$k = \frac{15 \pm \sqrt{(-15)^2 - 4 \times 2 \times 27}}{2 \times 2} = \frac{15 \pm \sqrt{225 - 216}}{4} = \frac{15 \pm 3}{4}$$ So, $$k_1 = \frac{15 + 3}{4} = \frac{18}{4} = 4.5$$ $$k_2 = \frac{15 - 3}{4} = \frac{12}{4} = 3$$ **Final answer:** The values of $k$ are $3$ and $4.5$ for the equation to represent a pair of straight lines.