Find the value of $k$ in order that the equation $\;$ $10x^2 + xy + ky^2 + x + 5y - 3 = 0$ $\;$ may represent a pair of straight lines.
Given equation is: $\;$ $10x^2 + xy + ky^2 + x + 5y - 3 = 0$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the general second degree equation
$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives
$a = 10, \; h = \dfrac{1}{2}, \; b = k, \; g = \dfrac{1}{2}, \; f = \dfrac{5}{2}, \; c = -3$
Equation $(2)$ represents a pair of straight lines if
$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ $\;\;\; \cdots \; (3)$
Substituting the values of $\;$ $a, \; b, \; c, \; f, \; g, \; h \;$ in equation $(3)$ gives
$10 \times k \times \left(-3\right) + 2 \times \dfrac{5}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} - 10 \times \left(\dfrac{5}{2}\right)^2 - k \times \left(\dfrac{1}{2}\right)^2 - \left(-3\right) \times \left(\dfrac{1}{2}\right)^2 = 0$
i.e. $\;$ $-30k - \dfrac{k}{4} + \dfrac{5}{4} - \dfrac{250}{4} + \dfrac{3}{4} = 0$
i.e. $\;$ $121 k = - 242$ $\implies$ $k = -2$