Author 
Topic 

AnnieD
New Member
Canada
2 Posts 
Posted  09/13/2007 : 18:37:10

I have a math assignment due tomorrow and there are a few questions that I don't even know where to begin, how to do, etc. Just overall stumpers. A group of 8 people from the class have been involved in an msn convo for the last 2hr trying to figure them out.. with no luck whatsoever. Any help/tips/start offs, etc. would be very much appreciated!
1) (3x5)(3x+1)² (3x+7) + 68 = 0
Answer: 1 ± √ 34 , 1 ± √ 2 *both these answers are over 3
2) (x²+6x+6)(x²+6x+8) = 528
Answer: 8, 2,  3 ± i√ 21
3) The height, length, and width of a small box are consecutive integers with the height being the smallest of the three dimensions. If the length and width are increased by 1cm each and the height is doubled, then the volume is increased by 120cm³. Find the dimensions of the original box.
Answer: 3cm, 4cm, 5cm



Subhotosh Khan
Advanced Member
USA
9117 Posts 
Posted  09/13/2007 : 20:23:12

quote: Originally posted by AnnieD
I have a math assignment due tomorrow and there are a few questions that I don't even know where to begin, how to do, etc. Just overall stumpers. A group of 8 people from the class have been involved in an msn convo for the last 2hr trying to figure them out.. with no luck whatsoever. Any help/tips/start offs, etc. would be very much appreciated!
1) (3x5)(3x+1)² (3x+7) + 68 = 0
Answer: 1 ± √ 34 , 1 ± √ 2 *both these answers are over 3
2) (x²+6x+6)(x²+6x+8) = 528
Answer: 8, 2,  3 ± i√ 21
3) The height, length, and width of a small box are consecutive integers with the height being the smallest of the three dimensions. If the length and width are increased by 1cm each and the height is doubled, then the volume is increased by 120cm³. Find the dimensions of the original box.
Answer: 3cm, 4cm, 5cm
We do not know where to start  unless you show some work/thoughts.
What have you been taught about solving polynomials, finding roots of polynomial, etc.?
Do you know calculus  do you know numerical analysis? 


skeeter
Advanced Member
USA
5634 Posts 
Posted  09/13/2007 : 20:51:44

1. use a substitution ...
let u = 3x+1
[(3x+1)6](3x+1)[(3x+1)+6] + 68 = 0
(u6)(u)(u+6) + 68 = 0
u(u36) + 68 = 0
u^{4}  36u + 68 = 0
(u2)(u34) = 0
u = 2, u = 34
(3x+1) = 2 , (3x+1) = 34
take it from here?
2. same technique ... let u = x+6x+6
u(u+2) = 528
u + 2u  528 = 0
(u  22)(u + 24) = 0
u = 22, u = 24
x+6x+6 = 22, x+6x+6 = 24
finish up.
3. three consecutive ... h, h+1, h+2
V = h(h+1)(h+2)
V + 120 = 2h(h+2)(h+3)
2h(h+2)(h+3)  h(h+1)(h+2) = 120
h(h+2)[2(h+3)  (h+1)] = 120
h(h+2)(h+5) = 120
you should be able to finish from here. 


skeeter
Advanced Member
USA
5634 Posts 
Posted  09/14/2007 : 13:16:58

your opinion is noted. 



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