a^2 - 2a + 1 /2 from a ^2 + 2a - 1 / 2

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- Dec 22nd 2009, 12:47 AMumamath
a^2 - 2a + 1 /2 from a ^2 + 2a - 1 / 2

- Dec 22nd 2009, 01:48 AMmr fantastic
- Dec 22nd 2009, 03:25 AMUnenlightened
I'm guessing

$\displaystyle a^2-2a+\frac{1}{2} - (a^2+2a-\frac{1}{2})$

You put all the $\displaystyle a^2$s together, all the parts with just $\displaystyle a$ together, and then all numbers with no $\displaystyle a$ together.

So you have

$\displaystyle a^2-a^2$= 0

$\displaystyle -2a - (2a)$ = -4a

$\displaystyle \frac{1}{2} - (-\frac{1}{2})$= 1

altogether giving

$\displaystyle -4a + 1$ - Dec 22nd 2009, 04:38 AMHallsofIvy
But he might well mean $\displaystyle \frac{x^2+ 2a+ 1}{2}- \frac{x^2- 2a+ 1}{2}$. That would be $\displaystyle \frac{x^2- x^2+-2a-(=2a)+ 1- 1}{2}$ and I will leave it to the OP to reduce that.