transforming formulas

**transgalactic** · October 8th 2011, 05:33 AM

i got my formula to this point
$q=(x_1+x_2)^2-x_2^2+(x_1+x_3)^2-2x_3^2$
how from this form i can get it to be with another form
in which we have $2(x_{1}+\frac{x_{2}}{2}+\frac{x_{3}}{2})^{2}$ member

?

**process91** · October 8th 2011, 05:41 AM

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

**transgalactic** · October 8th 2011, 05:42 AM

is there easiyer way?

**process91** · October 8th 2011, 05:43 AM

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

**Deveno** · October 8th 2011, 05:47 AM

are you still working on quadratic forms transgalactic?

**process91** · October 8th 2011, 05:48 AM

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

**transgalactic** · October 8th 2011, 06:30 AM

q= $2x_{1}^{2}+3x_{2}^{2}-x_{3}^{2}+2x_{1}x_{2}++2x_{1}x_{3}$
$=x_{1}^{2}+x_{1}^{2}+3x_{2}^{2}-x_{3}^{2}+2x_{1}x_{2}+2x_{1}x_{3}$
$=x_{1}^{2}+2x_{1}x_{2}+x_{1}^{2}+x_{2}^{2}+2x_{2}^ {2}-x_{3}^{2}+2x_{1}x_{3}=x_{1}^{2}+2x_{1}x_{2}+x_{1}^ {2}+x_{2}^{2}+2x_{2}^{2}-2x_{3}^{2}+x_{3}^{2}+2x_{1}x_{3}=$
$=(x_{1}+x_{2})^{2}+x_{1}^{2}+x_{2}^{2}+2x_{2}^{2}-2x_{3}^{2}+2x_{1}x_{3}+x_{3}^{2}=$

$=(x_{1}+x_{2})^{2}+x_{1}^{2}+x_{2}^{2}+2x_{2}^{2}-2x_{3}^{2}+(x_{1}+x_{3})^{2}-x_{1}^{2}=$

$=(x_{1}+x_{2})^{2}+3x_{2}^{2}-2x_{3}^{2}+(x_{1}+x_{3})^{2}$

in my solution they from the start show this member
$2(x_{1}+\frac{x_{2}}{2}+\frac{x_{3}}{2})^{2}$
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
?

**transgalactic** · October 8th 2011, 06:32 AM

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

**process91** · October 8th 2011, 07:15 AM

Oh, you just need to use the distributive law:

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

**transgalactic** · October 8th 2011, 08:45 AM

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

**process91** · October 8th 2011, 11:27 AM

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

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