Find the values of k for which [2x^2 + kxy + 3y^2 - 5y - 2]factors into two linear factors.

Find the values of k for which \(2x^2 + {\color{red}k}xy + 3y^2 - 5y - 2\) factors into two linear factor.

Source: https://www.askiitians.com/iit-jee-algebra/quadratic-equations-and-expressions/resolution-of-a-quadratic-function-into-linear-factors.aspx

In general:

\(\begin{array}{|lrcll|} \hline \text{Let } f(x, y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c \\ \text{writing this in descending powers of } y \text{ and equating to zero, we have } \\ by^2 + 2(hx + f)y + (ax^2 + 2gx + c) = 0 \\\\ \text{this is a quadratic equation in } y. \\ \text{Solving this for } y \text{, we get } y = \dfrac{ - (hx+f) \pm \sqrt{ (hx+f)^2 -b(ax^2+2gx+c)} } {b} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline \text{The quantity under radical must be perfect square, }\\ \text{which is quadratic in }x\text{ and for the desired result, its discriminant must be zero: } \\\\ (hx+f)^2 -b(ax^2+2gx+c) \\ = h^2x^2+2hfx+f^2-abx^2-2bgx-bc \\ = (h^2-ab)x^2+2(hf-bg)x+(f^2-bc) \\\\ x = \dfrac{-2(hf-bg)\pm \sqrt{(-2(hf-bg))^2-4(h^2-ab)(f^2-bc)} } {2(h^2-ab)} \\ x = \dfrac{-2(hf-bg)\pm \sqrt{(4(hf-bg)^2-4(h^2-ab)(f^2-bc)} } {2(h^2-ab)} \\ x = \dfrac{-2(hf-bg)\pm 2\sqrt{(hf-bg)^2- (h^2-ab)(f^2-bc)} } {2(h^2-ab)} \\\\ (hf-bg)^2- (h^2-ab)(f^2-bc) = 0 \\ h^2f^2-2hfbg+b^2g^2-h^2f^2+h^2bc+abf^2-ab^2c = 0 \\ -2hfbg+b^2g^2 +h^2bc+abf^2-ab^2c = 0 \quad | \quad : b \\ -2hfg+bg^2 +h^2c+af^2-abc = 0 \quad | \quad \cdot (-1) \\ 2hfg-bg^2 -h^2c-af^2+abc = 0 \\ \mathbf{abc+2hfg-af^2-bg^2-h^2c = 0} \quad | \text{ The condition of two linear factors } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline f(x, y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c \\ \hline 2x^2 + {\color{red}k}xy + 3y^2 - 5y - 2 \\ \text{Here, } a=2,\ 2h={\color{red}k},\ b=3, g=0,\ 2f = -5,\ c= -2 \\\\ \mathbf{abc+2hfg-af^2-bg^2-h^2c} &=& \mathbf{0} \\ 2\cdot 3\cdot(-2)+k\cdot\left( \dfrac{-5}{2} \right)\cdot 0 -2\cdot \left( \dfrac{-5}{2} \right)^2 - 3\cdot 0^2 -\left( \dfrac{k}{2} \right)^2\cdot(-2) &=& 0 \\ -12 -\dfrac{25}{2} +\dfrac{k^2}{2} &=& 0 \quad | \quad \cdot 2 \\ -24 - 25 + k^2 &=& 0 \\ -49 + k^2 &=& 0 \\ k^2 &=& 49 \\ \mathbf{k} &=& \mathbf{\pm 7} \\ \hline \end{array}\)

Result:

\(\begin{array}{|lrcll|} \hline k = 7: \\ & (2x+y-2)(x+3y+1)&=& 2x^2+6xy+2x+xy+3y^2+y-2x-6y-2 \\ &&=& 2x^2+{\color{red}7}xy+3y^2-5y-2 \\\\ k = -7: \\ & (2x-y+2)(x-3y-1)&=& 2x^2-6xy-2x-xy+3y^2+y+2x-6y-2 \\ &&=& 2x^2{\color{red}-7}xy+3y^2-5y-2 \\ \hline \end{array} \)