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 April 27th, 2013, 09:09 PM #1 Member   Joined: Mar 2013 Posts: 36 Thanks: 0 Algebra simplification Hi! I am unsure of the intermediate steps in these equations: 1. $\displaystyle -\frac13(x-3)^2+6$ $\frac13(x-3)^2= 6\,\Rightarrow\,(x-3)^2=18$ When I multiply through by -1, the equations don't remain equal. What's the intermediate step? 2. $\displaystyle \frac{-2\pm\sqrt{8}}{2}\,=\,-1\,\pm\,\sqrt{2}$ My simplification gives: $-1\,\pm\,2\sqrt{2}$
 April 28th, 2013, 04:28 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,957 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: Algebra simplification 1. If you mean $-\frac13(x\,-\,3)^2\,+\,6\,=\,0$, multiply through by 3. 2. $\frac{-2\,\pm\,\sqrt{8}}{2}\,=\,\frac{-2\,\pm2\sqrt{2}}{2}\,=\,-1\,\pm\sqrt{2}$
April 28th, 2013, 04:44 AM   #3
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Quote:
 Originally Posted by Aftermath When I multiply through by -1, the equations don't remain equal.
That doesn't make sense, as the equations (x - 3)² = 18 and -(x - 3)² = -18 are equivalent. Why would you want to multiply by -1 anyway? This problem started with an expression; should it have started with an equation?

April 28th, 2013, 02:41 PM   #4
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Re: Algebra simplification

Thank you.

Quote:
 Originally Posted by greg1313 1. If you mean $-\frac13(x\,-\,3)^2\,+\,6\,=\,0$, multiply through by 3.
$-(x-3)^2\,+\,18\,=\,0$

$-(x-3)^2\,=\,-18$

[/quote]2. $\frac{-2\,\pm\,\sqrt{8}}{2}\,=\,\frac{-2\,\pm2\sqrt{2}}{2}\,=\,-1\,\pm\sqrt{2}$[/quote]
I just realized that the two (2) is a shared denominator.

April 28th, 2013, 02:45 PM   #5
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Quote:
Originally Posted by skipjack
Quote:
 Originally Posted by Aftermath When I multiply through by -1, the equations don't remain equal.
That doesn't make sense, as the equations (x - 3)² = 18 and -(x - 3)² = -18 are equivalent. Why would you want to multiply by -1 anyway? This problem started with an expression; should it have started with an equation?
It doesn't make sense indeed, tired post.

I multiply by -1 to remove the negative before the parentheses:

$-(x-3)^2\,=\,-18$

$(x-3)^2\,=\,18$

By the way, why is it that multiplying through by -1 doesn't change the sign inside the parentheses:

$-(x-3)^2\,=\,-18$

$(x+3)^2\,=\,18$

 April 28th, 2013, 03:10 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,916 Thanks: 2199 The exponentiation would need to be done first, and then the multiplication by -1 could be done by changing the sign of each term within the parentheses, not just the second term within the parentheses. -(x - 3)² = -(x² - 6x + 9) = -x² + 6x - 9

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