partial fractions,,?

Question

i asked a different question a few minutes ago,,

http://answers.yahoo.com/question/index;_ylt=AjHskK2hhl2acPVDnBvayxsCxgt.?qid=20070528091854AAbUhmT

well heres my problem now

1\x(a-x) how do i wirte this in partial fractions,,

can it be done like this

A/x and B/(a-x) = 1\x(a-x)

now A(a-x) + Bx = 1

plug in x = 0 and find A which is 1/a and find b also like this

whats wrong in my method i, think i am wrong!!!

thanks

Answer 1

Yes first you need to write as
(A/x)+(B/a-x)=(1/x(x-a)
=>1=A(x-a) + Bx
=>if x=a then B=(1/a)
and substituting x=0 we get A = (-1/a)
Thus given equation can be written as (-1/ax)+(1/a(x-a))

Answer 2

Actually, your method is perfectly valid, and is usually the fastest way to determine the coefficients in partial fractions. The reason this works is because there is a theorem guaranteeing both the existence and uniqueness of the partial fraction decomposition, so if by plugging in coefficients you find that there is only one value for say, A, which could satisfy the relation for a particular value for x, the existence part of the theorem guarantees that there will be a partial fraction decomposition which works for all values of x, which then will have that value for A. N.B. -- this only works because we know ahead of time that there is a solution. If applied to a more general system of equations, it would only show that if there is a solution, that it must have that value for A, and additional work would be required to show that the values so obtained actually constitute a solution.

Answer 3

2x² + 5x - 12 = (2x - 3)(x + 4), 2 non-repeated linear factors So seek for a decomp of the shape eleven/(2x² + 5x - 12) = A/(2x - 3) + B/(x + 4). Multiply via via (2x - 3)(x + 4) to get eleven = A(x + 4) + B(2x - 3). There are a pair of techniques of finding A and B. With non-repeating linear factors, we are able to in user-friendly terms permit x be equivalent to the roots of the quadratic. if x = -4, you get eleven = B(-8-3) = -11B ==> B = -a million if x = 3/2, you get eleven = A(3/2 + 4) = (eleven/2)A ==> A = 2. The decomposition is for this reason eleven/(2x² + 5x - 12) = 2/(2x - 3) - a million/(x + 4). on the different hand, you need to evaluate coefficients on the left and precise. eleven = (A + 2B)x + (4A - 3B) giving A + 2B = 0, and 4A - 3B = eleven. you will get an identical constants A and B.

Answer 4

Up to the statement A(a-x) + Bx = 1 you are wright.

You should continue like this

Aa -Ax + Bx = 1

(B-A)x + Aa = 1

Then

B - A = 0

Aa = 1

This gives

A = B = 1/a

It means

1/x(a-x) = 1/ax + 1/a(a-x)

Answer 5

Your concern is "think i am wrong!!!"

When you get an answer you worry about, CHECK IT!
Your method produced A = B= (1/a)

so (1/a) / x + (1/a) / (a-x) = 1/ax + 1/[a(a-x)] cross multiply

=( [a(a-x)] + ax) / [(ax)(a-x)]

= a^2x -ax + ax) / [a^2(x)(a-x)]

= a^2 / [a^2(x)(a-x)]

= 1 / [(x)(a-x)]

so you proved your answer results in the expression you started with and you can be 100% confident that you are correct!

Good work!

Answer 6

See this site for a thorough explanation: too long for here:
http://www4.ncsu.edu/unity/lockers/users/f/felder/public/kenny/papers/partial.html

Answer 7

All of your work seems to be correct and make sense to me. Perhaps you are correct and don't even know it! :)