# Showing that the following are equivalent

1. Feb 8, 2013

### SithsNGiggles

1. The problem statement, all variables and given/known data

I had this posted under a different question a while back and didn't get any responses, so I thought I'd rephrase it. I've reduced it to what I think I'm supposed to show. (Here's the old post: https://www.physicsforums.com/showthread.php?p=4260206#post4260206 ... Disregard the actual coefficients, I fudged those a bit.)

2. Relevant equations

3. The attempt at a solution
I have to show the following:
$||x||^2 = \frac{2}{3}|x_1|^2+\frac{1}{6}|x_1-\sqrt3x_2|^2+\frac{1}{6}|x_1+\sqrt3x_2|^2$

$x\in\mathbb{R}^2$, so $x=(x_1,x_2)$, and the left side can be rewritten so that I have
$|x_1|^2+|x_2|^2 = \frac{2}{3}|x_1|^2+\frac{1}{6}|x_1-\sqrt3x_2|^2+\frac{1}{6}|x_1+\sqrt3x_2|^2$

Is there any way to do this?

2. Feb 8, 2013

### SammyS

Staff Emeritus
|a + b|2 = (a + b)2

The rest is pretty basic algebra.

Merely expand the right hand side & collect terms.

3. Feb 8, 2013

### SithsNGiggles

Is it really that simple? I was getting the impression I couldn't do that. I thought for some reason that the sign of x1 or x2 would make a difference. Thanks a lot!

4. Feb 9, 2013

### SammyS

Staff Emeritus
If a+b ≥ 0 , then it's pretty obvious that |a+b|2 = (a+b)2 , since |a+b| = a+b .

If a+b < 0 , then |a+b| = -(a+b), so that |a+b|2 = (-(a+b))2. But that's obviously equal to (a+b)2.

5. Feb 9, 2013

### SithsNGiggles

Oh I see it now, thanks! I was thinking that the sign of x1 and x2 would have some effect on the sum. Thanks again.