i got my formula to this point
$\displaystyle q=(x_1+x_2)^2-x_2^2+(x_1+x_3)^2-2x_3^2$
how from this form i can get it to be with another form
in which we have $\displaystyle 2(x_{1}+\frac{x_{2}}{2}+\frac{x_{3}}{2})^{2}$ member
?
q=$\displaystyle 2x_{1}^{2}+3x_{2}^{2}-x_{3}^{2}+2x_{1}x_{2}++2x_{1}x_{3}$
$\displaystyle =x_{1}^{2}+x_{1}^{2}+3x_{2}^{2}-x_{3}^{2}+2x_{1}x_{2}+2x_{1}x_{3}$
$\displaystyle =x_{1}^{2}+2x_{1}x_{2}+x_{1}^{2}+x_{2}^{2}+2x_{2}^ {2}-x_{3}^{2}+2x_{1}x_{3}=x_{1}^{2}+2x_{1}x_{2}+x_{1}^ {2}+x_{2}^{2}+2x_{2}^{2}-2x_{3}^{2}+x_{3}^{2}+2x_{1}x_{3}=$
$\displaystyle =(x_{1}+x_{2})^{2}+x_{1}^{2}+x_{2}^{2}+2x_{2}^{2}-2x_{3}^{2}+2x_{1}x_{3}+x_{3}^{2}=$
$\displaystyle =(x_{1}+x_{2})^{2}+x_{1}^{2}+x_{2}^{2}+2x_{2}^{2}-2x_{3}^{2}+(x_{1}+x_{3})^{2}-x_{1}^{2}=$
$\displaystyle =(x_{1}+x_{2})^{2}+3x_{2}^{2}-2x_{3}^{2}+(x_{1}+x_{3})^{2}$
in my solution they from the start show this member
$\displaystyle 2(x_{1}+\frac{x_{2}}{2}+\frac{x_{3}}{2})^{2}$
but i am not used to work with its formula
i prefer the simple (a+b)^2 form
you said that i should dissasmble every colse on both expressions and regroup
but in the test i idont know the other expression
so how to get this member into my formula
?