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AnnieD
New Member

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) (3x-5)(3x+1)² (3x+7) + 68 = 0Answer: -1 ± √ 34 , -1 ± √ 2*both these answers are over 32) (x²+6x+6)(x²+6x+8) = 528Answer: -8, 2, - 3 ± i√ 213) 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

USA
9117 Posts

 Posted - 09/13/2007 :  20:23:12 quote:Originally posted by AnnieDI 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) (3x-5)(3x+1)² (3x+7) + 68 = 0Answer: -1 ± √ 34 , -1 ± √ 2*both these answers are over 32) (x²+6x+6)(x²+6x+8) = 528Answer: -8, 2, - 3 ± i√ 213) 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, 5cmWe 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

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(u-6)(u)(u+6) + 68 = 0u(u-36) + 68 = 0u4 - 36u + 68 = 0(u-2)(u-34) = 0u = 2, u = 34(3x+1) = 2 , (3x+1) = 34take it from here?2. same technique ... let u = x+6x+6u(u+2) = 528u + 2u - 528 = 0(u - 22)(u + 24) = 0u = 22, u = -24x+6x+6 = 22, x+6x+6 = -24finish up.3. three consecutive ... h, h+1, h+2V = h(h+1)(h+2)V + 120 = 2h(h+2)(h+3)2h(h+2)(h+3) - h(h+1)(h+2) = 120h(h+2)[2(h+3) - (h+1)] = 120h(h+2)(h+5) = 120you should be able to finish from here.

skeeter

USA
5634 Posts

 Posted - 09/14/2007 :  13:16:58 your opinion is noted.
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