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August 1st, 2015, 11:38 AM   #1
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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?

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August 1st, 2015, 01:42 PM   #2
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Check the derivation of your
equation - 10 is too large.
The volume = LWH =
(40 - 4)(32 - 4)·2 = 2016.
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August 1st, 2015, 03:40 PM   #3
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Quote:
Originally Posted by uvkajed View Post
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?
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August 1st, 2015, 04:13 PM   #4
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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.
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August 1st, 2015, 08:34 PM   #5
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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
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August 1st, 2015, 10:35 PM   #6
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Quote:
Originally Posted by Azzajazz View Post
I think all three values of $x$ are answers to the question.
The value 23 (approximately) is too large - the corners would overlap.
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August 1st, 2015, 10:58 PM   #7
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Quote:
Originally Posted by skipjack View Post
The value 23 (approximately) is too large - the corners would overlap.
Yeah. Brain fail. Thanks for the correction.
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