Expressing a quadratic form in canonical form using Lagrange

[Apothem]

Problem:
Express the quadratic form:

q=x₁x₂+x₁x₃+x₂x₃

in canonical form using Lagrange's Method/Algorithm

Attempt:
Not really applicable in this case due to the nature of my question

The answer is as follows:
Using the change of variables:
x₁=y₁+y₂
x₂=y₁-y₂
x₃=y₃
Indeed you get q'=(y₁+y₃)²-(y₂)²-(y₃)²

I can see that when you substitute it in it does indeed reduce it to canonical form however my question is as follows:
How am I meant to arrive at the above change of variables? I have seen a few cases of completing the square, but to my knowledge we can't use that here. EDIT: We can

Any help is greatly appreciated!

EDIT:
I think I've been able to deduce it...
x₁x₂+x₁x₃+x₂x₃=(x₃)²+x₃(x₂+x₁)-(x₃)² and from there I completed the square, the only thing I'm still a bit uncertain on is that I worked this out the other way i.e. I worked out y in terms of x and not x in terms of y...

 

[Asim Riaz]

Is this ok? Yes, that is fine. Completing the square is a valid approach to solving this problem and it is essentially the same as Lagrange's Method/Algorithm. By completing the square, you are taking a quadratic expression of the form ax2 + bx + c and rewriting it as (x - x0)2 + d, where x0 is the vertex of the parabola. That is essentially what you are doing in your solution, and it is equivalent to using Lagrange's method.