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 August 1st, 2015, 12:38 PM #1 Member   Joined: Sep 2014 From: Toronto Posts: 43 Thanks: 0 Need help asap! Hey guys! I'm having trouble with the question is listed below. Any help would be appreciated. 4x^3-144x^2-1280x=2016, what's the correct value for x?
 August 1st, 2015, 02:42 PM #2 Senior Member   Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 Check the derivation of your equation - 10 is too large. The volume = LWH = (40 - 4)(32 - 4)·2 = 2016.
August 1st, 2015, 04:40 PM   #3
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Quote:
 Originally Posted by uvkajed Check the derivation of your equation - 10 is too large. The volume = LWH = (40 - 4)(32 - 4)·2 = 2016.
What do you mean by derivation of my equation? Also how'd you get 2 as the final answer?

 August 1st, 2015, 05:13 PM #4 Math Team   Joined: Nov 2014 From: Australia Posts: 672 Thanks: 239 Everything I can see is correct except for the line $x = \dfrac{34\pm\sqrt{148}}{2}\implies x = 10$ or $x = 23$. I suggest you try this calculation again. You got an answer of 2 as well. It's at the end of the section marked $(1)$. To answer your original question, I think all three values of $x$ are answers to the question.
 August 1st, 2015, 09:34 PM #5 Senior Member   Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 Actually, it wasn't hard to guess an answer, but... (40 - 2x)(32 - 2x)x = 2016, or (20 - x)(16 - x)x = 504, from which x^3 - 36x^2 + 320x - 504 = 0. x must be even (consider parity). Let x = 2x'; then 8x'^3 - 144x'^2 + 640x' - 504 = 0, or x'^3 - 18x'^2 + 80x' - 63 = 0. By inspection, x' = 1. Then 1 - 18 + 80 - 63 = 0 checks. Then x = 2x' = 2. I didn't bother checking any further. Thanks from aurel5
August 1st, 2015, 11:35 PM   #6
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Quote:
 Originally Posted by Azzajazz I think all three values of $x$ are answers to the question.
The value 23 (approximately) is too large - the corners would overlap.

August 1st, 2015, 11:58 PM   #7
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Quote:
 Originally Posted by skipjack The value 23 (approximately) is too large - the corners would overlap.
Yeah. Brain fail. Thanks for the correction.

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