In the following question I have to present the bilinear form as sum of squares with Lagrange method.
$$q(x_1,x_2,x_3,x_4)=2x_1x_4-6x_2x_3$$
However I don't know how I can do it here since none of the coefficients here is squared. I have seen the following solution, however I don't know how to reach it:
$$q=2x_1x_4-6x_2x_3=\frac{1}{2}(x_1+x_4)^2-\frac{1}{2}(x_1-x_4)^2+\frac{3}{2}(x_2-x_3)^2-\frac{3}{2}(x_2-x_3)^2$$
I don't understand the logic here.
What are the formulas that were used to "complete the square" here?
Thanks,
Alan
Difference of two squares: $$ a^2-b^2=(a+b)(a-b). $$ Now put $a=x+y$ and $b= x-y$. Then $a+b=2x$, $a-b=2y$, and $$ (x+y)^2-(x-y)^2 = 4xy $$