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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

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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 $$ (x+y)^2-(x-y)^2 = 4xy $$

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  • $\begingroup$ Thank you! It's quite clear. $\endgroup$ – Alan Oct 3 '15 at 12:05

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