Thread: transforming formulas

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

?

Just multiply out both forms, then put them back together in pieces.

is there easiyer way?

It takes about 5 minutes. I just did it in less time than that.

are you still working on quadratic forms transgalactic?

The resulting formula doesn't appear that "nice", perhaps you could post the context in which you arrived at this question.

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
?

Originally Posted by Deveno

are you still working on quadratic forms transgalactic?

i am learning to a test so i try to solve every sort of question

Oh, you just need to use the distributive law:

$\displaystyle (a+b+c)^2=(a+b+c)(a+b+c)=a^2 +ab+ac+ba+b^2+bc+ca+cb+c^2=a^2+b^2+c^2+2(ab+bc+ac)$

its not what i am asking.
i want the from this form
$\displaystyle =(x_{1}+x_{2})^{2}+3x_{2}^{2}-2x_{3}^{2}+(x_{1}+x_{3})^{2}$
ill get
(a+b+c)^2 member

That's not the same as your first post, so I'm afraid I'm unsure of what you're asking.