How to factorize equation of pair of lines?

Question

Find two equations of lines : $$6x^2 -xy -12y^2 -8x + 29y -14= 0$$

I tried doing it using quadratic equation but got some very much unusual value. How to solve this?

Answer 1

Expand and collect:

$$(ax+by+c)(dx+ey+f)=(ad)\ x^2 + (ae+bd)\ xy + (be)\ y^2 + (af+cd)\ x + (bf+ce)\ y + cf$$

Identifying coefficients leads to solving this system:

$\begin{cases} ad=6\\ ae+bd=-1\\ be=-12\\ af+cd=-8\\ bf+ce=29\\ cf=-14\end{cases}$

You can try to work out integer solutions, since this is equal to $0$ there will be a proportionality coefficient, for instance try to start with $a=2,d=3$ or $c=2,f=-7$.

This is what I found: $$(3x+4y-7)(2x-3y+2)=0$$

Answer 2

Hint:

$$6x^2-x(y+8)-(12y^2-29y+14)=0$$

The discriminant $(D)$ is $$(y+8)^2+24(12y^2-29y+14)=(17y-20)^2$$

$$\implies x=\dfrac{y+8\pm\sqrt D}{2\cdot6}$$

Answer 3

you can also partially diff. the equation F(x,y)=` 6x2−xy−12y2−8x+29y−14 strong text dF/dx=0 ...(i) dF/dy=0 ...(ii) (i) and(ii) will give unique linear equations.