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August 1st, 2015, 11:38 AM  #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^3144x^21280x=2016, what's the correct value for x? 
August 1st, 2015, 01:42 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, 03:40 PM  #3 
Member Joined: Sep 2014 From: Toronto Posts: 43 Thanks: 0  
August 1st, 2015, 04:13 PM  #4 
Math Team Joined: Nov 2014 From: Australia Posts: 686 Thanks: 243 
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, 08:34 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. 
August 1st, 2015, 10:35 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,841 Thanks: 1564  
August 1st, 2015, 10:58 PM  #7 
Math Team Joined: Nov 2014 From: Australia Posts: 686 Thanks: 243  

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