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April 8th, 2017, 11:19 AM   #11
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Quote:
Originally Posted by skeeter View Post
$-x^2-6x+2$

$-(x^2+6x-2)$

$-\left[(x^2+6x+9) - 9 - 2\right]$

$-\left[(x+3)^2 - 11\right]$

$-(x+3)^2 + 11$
Oh do you not do this:
$\displaystyle -[(x+3)^2-9]-2$
and why not?
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April 8th, 2017, 11:33 AM   #12
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Quote:
Originally Posted by jamesbrown View Post
Oh do you not do this:
$\displaystyle -[(x+3)^2-9]-2$
and why not?
(1) because that is incorrect

$-[(x+3)^2-9]-2 = -(x+3)^2 + 9 - 2 = -(x+3)^2 + 7 \ne -(x+3)^2 + 11$


(2) because I prefer putting everything inside $-[bracket]$.

if you want to leave the +2 outside the $-[bracket]$ ...

$2 - 6x - x^2 = -x^2 - 6x + 2$

$-(x^2 + 6x) + 2$

$-(x^2 + 6x + 9 - 9) + 2$

$-\left[(x^2 + 6x + 9) - 9\right] + 2$

$-\left[(x+3)^2 - 9\right] + 2$

$-(x+3)^2 + 9 + 2$

$-(x+3)^2 + 11$
Thanks from jamesbrown
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April 8th, 2017, 12:28 PM   #13
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Quote:
Originally Posted by skeeter View Post
(1) because that is incorrect

$-[(x+3)^2-9]-2 = -(x+3)^2 + 9 - 2 = -(x+3)^2 + 7 \ne -(x+3)^2 + 11$


(2) because I prefer putting everything inside $-[bracket]$.

if you want to leave the +2 outside the $-[bracket]$ ...

$2 - 6x - x^2 = -x^2 - 6x + 2$

$-(x^2 + 6x) + 2$

$-(x^2 + 6x + 9 - 9) + 2$

$-\left[(x^2 + 6x + 9) - 9\right] + 2$

$-\left[(x+3)^2 - 9\right] + 2$

$-(x+3)^2 + 9 + 2$

$-(x+3)^2 + 11$
That makes a lot more sense. What I was doing, was putting inside the brackets and then out of the brackets again. Without changing the sign.

Thanks for all the help
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