If $3x^2+2\alpha xy+2y^2+2ax-4y+1$ can be resolved into two linear factors, prove that $\alpha$ is the root of the equation $x^2+4ax+2a^2+6=0$.

Question

If $3x^2+2\alpha xy+2y^2+2ax-4y+1$ can be resolved into two linear factors, prove that $\alpha$ is the root of the equation $x^2+4ax+2a^2+6=0$.

I know this question has already been asked on Math.SE here, but is there a way to solve this by only using the concepts of quadratic equations? I am not very clear with the solution given there.

Answer 1

The discriminant of the equation $p(x,y) = 3x^2+(2\alpha y+2a)x +2y^2-4y+1=0$ is $\delta = (2\alpha y+2a)^2-12(2y^2+4y+1)$.

Thus, $$p(x,y) = 3\left(x-\frac{-(2\alpha y+a) +\sqrt{\delta}}{6}\right)\left(x-\frac{-(2\alpha y+a)-\sqrt{\delta}}{6}\right)$$

Now $p(x,y)$ is a perfect square iff the discriminant $\delta_1$ of $\delta =0$ is $0$. We can verify $\delta_1 =0 \Leftrightarrow \cdots \Leftrightarrow \alpha^{2}+4\alpha a+ 2a^2+6=0$. Equivalently, $\alpha$ is a root of $x^2+4ax+2a^2+6=0$.