Anyone any good at maths?

Jon

Jon

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I have no idea how to resolve a partial fraction with a quadratic???

I.e

6x-5
--------------
(x-4)(x^2+3)

ANY help would be appreciated, as you can probably tell i had no idea where to post this..:blink:
 
I actually looked at the wiki page and i've missed the section devoted to the quadratic factor.. It explains it quite well but i must admit (this may be because i've done several 12 hour shifts in a row) it's still going over my head.. but thanks anyway.
 
that fraction should be equal to what?

if zero, than the "upper part" of the fraction has to be zero - so its easy,
if something else, you have to turn it into this form: ax?+bx+c=0 or better x?+(b/a)x+(c)=x?+px+q=0
then x1=-(p/2)+sqrt((p/2)?-q) and x2=-(p/2)-sqrt((p/2)?-q)
 
Last edited:
I do not understand the question. What do you want to solve it to?
 
Just a bit about the partial fraction that was mentioned

Sorry it's difficult to express for me because i don't know the English words.

In your denominator you have a polynom of the 3rd degree.
It's already partially impanded, means in the form of (x-x1)*(x-x2)*(x-x3) where x1, x2, x3 are the roots.

(x-4)(x^2+3)=0

you can clearly see that x=4 must be a root.

for (x^2+3)=0 you get imaginary solutions

x2=j*sqrt(3) and x3=-j*sqrt(3)

When you have a real solution and simple complex ones (means that there is no ^2 or ^3 or ^4... over the this bracket), you must stick to that case in the wikipedia article:

http://en.wikipedia.org/wiki/Partial_fraction#An_irreducible_quadratic_factor_in_the_denominator

You have to follow that scheme in the example. I tried it on paper, but used my formula book where they used different letters.

A in Wikipedia is A in my example
B in Wikipedia is P in my example
C in Wikipedia is Q in my example

http://img91.imageshack.**/img91/6812/cimg8153tb8.jpg
http://img87.imageshack.**/img87/2447/cimg8154ur6.jpg


I'm not sure if you really need to do that. But that is THE method to use if u want to integrate such an expression
 
Moved from SITE HELP, to Off-topic... this isn't bittorrent or video related.
 
What do you want to do with it? If you just expand the denominator it should be simplified. Is there a specific form you're trying to reach?
 
idk said:
Just a bit about the partial fraction that was mentioned

Sorry it's difficult to express for me because i don't know the English words.

In your denominator you have a polynom of the 3rd degree.
It's already partially impanded, means in the form of (x-x1)*(x-x2)*(x-x3) where x1, x2, x3 are the roots.

(x-4)(x^2+3)=0

you can clearly see that x=4 must be a root.

for (x^2+3)=0 you get imaginary solutions

x2=j*sqrt(3) and x3=-j*sqrt(3)

When you have a real solution and simple complex ones (means that there is no ^2 or ^3 or ^4... over the this bracket), you must stick to that case in the wikipedia article:

http://en.wikipedia.org/wiki/Partial_fraction#An_irreducible_quadratic_factor_in_the_denominator

You have to follow that scheme in the example. I tried it on paper, but used my formula book where they used different letters.

A in Wikipedia is A in my example
B in Wikipedia is P in my example
C in Wikipedia is Q in my example

http://img91.imageshack.**/img91/6812/cimg8153tb8.jpg
http://img87.imageshack.**/img87/2447/cimg8154ur6.jpg


I'm not sure if you really need to do that. But that is THE method to use if u want to integrate such an expression
Click to expand...

Oh man... we've just studied this method of integration last week at school. :| Sometimes the length of a single integration is staggering.
 
vikiradTG2007 said:
Oh man... we've just studied this method of integration last week at school. :| Sometimes the length of a single integration is staggering.
Click to expand...

thevictor390 said:
^ Wait until you get to double and triple integrals
Click to expand...

Bahhhh!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!


Has nothing to do with the original question... but reminds me of this which was really... annoying until it really got a correct result :cry:

http://img440.imageshack.**/img440/1328/cimg8155yi7.jpg
http://img509.imageshack.**/img509/1729/cimg8156rw7.jpg
 
Oh.

My.

God.




I need two cans of Red Bull before attempting a single integral of that kind.


Just out of curiosity, when did you start studying integrals? High-school or university?
 
Basic and simple ones in school. Like all the distance-time-acceleration relations. And Integral of a force along a distance = energy. Also area between 2 functions etc

But double and triple at university. But right at the beginning there, just after school. And not much explanation. We got just an exercise and had to do it for a specific date. Was quiet hard to find out everything by ourselves for the first time.
 
So I gather that you're learning Maths and Physics in a tightly-linked sort of way in high school, is that right? Over here, maybe also because of the teachers, they seem broken completely apart in terms of method of teaching and of applications of one in the other.
 
As for physics and maths yes. But you couldn't claim that for any other subjects.
 
If you're getting an engineering or physics degree, wait until you start mixing linear algebra and calculus...

Fourier transforms & hilbert spaces & manifolds, oh my!
 
So did TC's question ever get answered? I'm not sure what he was asking.
 
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