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If we are given a general quadratic function in $x, y$, there is a condition that it can be resolved into two linear factors of the form $ax+by+c$. I found the following proof for this.

If we have a quadratic function

$$f(x, y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c$$

To factorize this, it is sufficient to find the roots of the equation:

$$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$

Considering this to be a quadratic in $x$ and applying the quadratic formula yields

$$x = \frac{-(hy + g) \pm \sqrt{(hy+g)^2 - a(by^2 + 2fy + c)}}{a}$$

$$\implies ax + hy + g = \pm \sqrt{(hy+g)^2 - a(by^2 + 2fy + c)}$$

Now, it is claimed that if $f(x, y)$ is the product of two linear factors the quantity under the square root sign must be a perfect square ie. the expression must be resolvable in to two identical linear factors.

As the expression under the square root is a quadratic in $y$ we can set its discriminant equal to zero and get the condition.

However, why must the expression be a perfect square if the quadratic has two linear factors? I was not able to exactly understand the reason behind it.

If we can write

$$f(x, y) = (px + qy + r)(p'x + q'y + r') = 0$$

Then

$$px + qy + r = 0 \implies ax + hy + g = (a-p)x + (h-q)y + (g-r)$$

which should mean that the quantity under the radical is a perfect square. However, I am not sure if this is the correct reason. Moreover, how do we go the other way, I have proved that if the quadratic is resolvable into two linear factors then the quantity under the square root is a perfect square. How do I prove the converse?

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Going with your last paragraph (with a notation change for readability) suppose $f(x,y)=(px+qy+r)(sx+ty+u).$ Multiplying things out and fitting it into $ax^2 + 2hxy + by^2 + 2gx + 2fy + c$ we have $$a=ps,\ h=(pt+qs)/2,\ b=qt,\\ g=(pu+rs)/2,\ f=(qu+rt)/2,\ c=ru.$$ Then (with the aid of some algebra software!) it turns out that $$(hy+g)^2-a(by^2+2fy+c)=(1/4)(rs-pu+(qs-pt)y)^2.$$ So the discriminant under the radical is automatically a square when $f$ factors into two linear terms. (This assumes $a \neq 0$ so we have a genuine quadratic in $x$ at the start.)

I haven't looked yet at going from discriminant a square backwards to the linear factorization. Maybe someone else can do that, but I'll try for something soon.

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