Who is good at math?

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Re: Who is good at math?

Postby Helaku on Wed Jun 16, 2010 3:19 pm

Turky wrote:15/5 is 3, not 5, that's where I went wrong. This tends to happen when simple arithmetic is the least of your problems.


amen.
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Re: Who is good at math?

Postby Magyk on Wed Jun 23, 2010 11:17 am

I could really yews some help on this particular problem; I'm stumped. :(
Solving for X...
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Re: Who is good at math?

Postby Maximus on Wed Jun 23, 2010 11:29 am

Multiply both sides by (2x^2-8x) to eliminate the denominator.
Combine like terms by adding 2x^2 and -8x to both sides (algebraic addition, where adding a negative is the same as subtracting a positive).
Factor out the common terms leaving 2x*(x^2-16)=0.
You're dealing with a trinomial which will necessarily have three solutions for x; either one real and two imaginary, or three real.
By that theorem I forget the name of, you set the individual terms equal to 0 and solve. The x^2-16=0 I hope you recognize is difference of squares.
You should be able to get the answer from this point.
I left out a lot intentionally; from 5 years tutoring math (mostly algebra and stats, some calculus), I know that doing all the work for someone doesn't really teach them anything. If you want more assistance, ask me some time when I'm not at work...
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Re: Who is good at math?

Postby Magyk on Wed Jun 23, 2010 11:37 am

That actually really helped a lot, I got 0,4,-4 for my answers.. would those be correct?

Also; I have one last problem that I'm having a hard time with and then I'm finally done. I drew out the picture in paint so you could see the problem.

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Re: Who is good at math?

Postby Helaku on Wed Jun 23, 2010 11:40 am

Multiply both sides by (2x^2-8x) to eliminate the denominator.

2x^3 -2x^2 -24x= -2x^2 +8x
2x^3 -32x=0
x( 2x^2 -32)=0
x=0 V 2x^2 -32=0
2x^2=32
x^2=16
x=4 V x= -4


There we go, the full procedure.
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Re: Who is good at math?

Postby Turky on Wed Jun 23, 2010 11:41 am

dude trying to redeem himself with simple maths
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Re: Who is good at math?

Postby Helaku on Wed Jun 23, 2010 11:55 am

v=a * b * c

a=x+4
b= this is the one we're looking for
c= x+1

a* c * b = v ( commutation when multiplying )

Since you have ( x + 1 ) in the denominator, you want to multiply both sides with it.

That will give you ( x + 4 ) * ( x + 1 )^2 * b = x^3 + 7x^2 +14x + 8

From there, put the (x+1) to second power, and then multiply it by (x+4), while rewriting the right side.

You should get ( x^3 + 6x^2 + 9x + 4 ) * b = x^3 + 7x^2 + 14x +8

From there,

x^3+7x^2+14x+8
b = ----------------------
x^3 +6x^2+ 9x + 4


A simple polynomial division and you're done.
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Re: Who is good at math?

Postby Helaku on Wed Jun 23, 2010 11:58 am

Turky wrote:dude trying to redeem himself with simple maths



I know as much math as you do, it's just that I don't speak math in english, so when you say fractions, I have to check a dictionary and find out that is my 5th grade elem math :P
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Re: Who is good at math?

Postby tens0r on Wed Jun 23, 2010 3:16 pm

Helaku wrote:
Turky wrote:dude trying to redeem himself with simple maths



I know as much math as you do, it's just that I don't speak math in english, so when you say fractions, I have to check a dictionary and find out that is my 5th grade elem math :P

Because fractions are only used in 5th grade. Much like numbers are only used in kindergarden.
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Re: Who is good at math?

Postby Turky on Wed Jun 23, 2010 3:43 pm

Helaku wrote:
Turky wrote:dude trying to redeem himself with simple maths



I know as much math as you do


As much as I do?

I hold a B.S. in Physics with a minor in math... not to even mention linear and abstract algebra, that's a whole different animal from classical algebra and calculus.
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