Question

If $2x^2+7xy+3y^2+8x+14y+\lambda =0$ can be resolved into $2$ linear factors, then the value of $\lambda$ is

Answer 1

The equation $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$
can be resolved into $2$ linear factors, if
$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

Hence, $2x^2+7xy+3y^2+8x+14y+\lambda =0$ can be resolved into $2$ linear factors, if
$6\lambda +2\left(7\right)\left(4\right)\left(\frac{7}{2}\right)-2(7{)}^2-3(4{)}^2-\lambda {\left(\frac{7}{2}\right)}^2=0\Rightarrow 6\lambda +196-98-48-\frac{49\lambda }{4}=0\Rightarrow -\frac{25\lambda }{4}+50=0\Rightarrow \lambda =\frac{50\times 4}{25}=8$