A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
can someone help me with partial fraction ? 4/x(x^2+4)
anonymous
 5 years ago
can someone help me with partial fraction ? 4/x(x^2+4)

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the function is 4/( x * ( x^2 + 4) )?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cool, so what's the first step? separate the function based upon the denominator, you have two functions mutliplied by eachother on the bottom: x * (x^2+4)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you could create a sum of two fractions: one with x as the denonm. and one with (x^2+4) as the denom. can you post those please?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i don't quite get it.. ><

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0np, ok, the goal is to express the function as the sum of other functions, which are actually "fractions"...where the denominators of these functions that you want to sum are the parts of the denominator of the larger function....clear?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so basically, here you have 4 ?( x * (x^2 + 4) )...the larger function on the denom is (x * (x^2 + 4) )....so can you separate this into two functions?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.04 / x + 1/(x^2+4) im not very sure.. i suck at maths

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right now, ignore the numerator so you're right, but ignore 4...the setup is now: \[4/(x * (x^2 + 4) ) = A/(x) + B/(x^2+4)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0does that look familiar, like something in your text?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but shouldn't it be A / x + Bx+c / (x^2+4)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, why do you think that you use "x" twice?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because of the demo has a (x^2)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, you are right about that equation not being quite right, because there is x^2+4...so you actually need to adjust the numerator of the fraction with (x^2+4)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0good catch! the equation becomes: \[4/(x∗(x2+4))=A/(x)+(Bx+C)/(x2+4) \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeap.. that about it.. the only thing is i don't know how to solve for Bx+C

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, now do you think you can make these fractions so that they have a common denominator? hint: multiply by "1"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i did a cross multipy which end up to be 4= A(x^2+4) + X(Bx+c)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, however you are adding these fractions...so you cannot cross multiply...you need to "give" the fractions the same denominator...and easy way would be multiply each fraction by 1, where 1 is actually a fraction with the numerator and denominator equal, in this case you would want to multiply A/x by (x^2+4)/(x^2+4)...see how this multiplying by 1 is OK?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the fraction on becomes: \[( A * (x^2 + 4))/(x * (x^2 +4))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nice, now can you apply the concept to the other fraction? then post the new equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A(x^2+4) / x(x^2+4) + Bx+C / x(x^2+4) am i right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0almost, what about the numerator on the second fraction?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the second fraction demo is x(x^2+4) isnt it same as the 1st fraction 's demo?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, that's right, but the numerator is missing something...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(A * (x^2 + 4) ) /( x * x^2 +4 )) + ((Bx + C)*(????))/(x * (x^2 + 4))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right now you've only multiplied the second fraction by (1/x)...so you didn't multiply by "1" like you did for the first with (x^2+4)/(X^2+4)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then is it Bx+C * x(x^2+4)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you're close, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i got to muplity Bx+C with X(x^2+4)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not quite, right it is multiplied by 1/x...see how you put x in the denominator? multiply the numerator by x too...so that you have multiplied by "1"..right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Bx+c * 1/x / x(x^2+4) ?? im confused

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, you were very close with (Bx + c) /x * (x^2 + 4)...how did you get here from (Bx+c)/(x^2 + 4)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh cancel out the x from the x(9x^2+4) but then the 1st fraction's demo isnt the same as 2nd one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually do the opposite....so (Bx+c)/(x^2+4) becomes ((Bx+c)*x)/((x^2+4)*x)...see how you multiplied numerator and denominator by x because you mutliplied the fraction by "1" where "1" = x/x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so, what are the two fractions now, now that they have common denominators of x*(x^2+4)...rewrite the equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A( x^2+4) / x(x^2+4) + Bx+c * (x) / x (x^2+4)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the entire equation looks like this: \[4/(x * (x^2 + 4)) = ((A * (x^2 + 4)) + ((Bx + c) *(x)))/(x * (x^2 + 4))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nice, now you can ignore denominators for the time being...so we actually want to look at the equality between the numerators

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[4 = (A *(x^2 + 4)) + (Bx+c)*x \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this eqn isnt it same as what just now i had cross multiply them together?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm not sure what you mean by cross multiply...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as stated just now A/x + Bx+c/ x^2+4 i cross mutiply them.. meaning. A * (x^2+4) and x* (bx+c)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in the original equation you're saying that you multiplied everything by the denominator?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cross multiply refers to when you have two fractions: like a/b = c/d , cross multiplying would yield a*d = c*b...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is what crossmultiplying technically is...it is best to the procedure as above so as to not move too hastily and miss a step...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so anyway, we are close to solving

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.04=(A∗(x2+4))+(Bx+c)∗x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[4 = (A * (x^2 + 4)) +(Bx + c)*(x)\] expand this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.04 = Ax^2 + 4A + Bx^2 + cx?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[4 = (A * (x^2 + 4)) + (Bx + C) * x = Ax^2 + 4A + Bx^2 + Cx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so, now, let's consider this, both equations are assumed to be equal for all values of x...right, because it's an equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you see how only one term does not have x in it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right on! so now we need to figure out what each letter should equal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, the simplest case was to consider that x = 0, now we have that A =1...so let's find C

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there are no x terms on the left side, so, what do you think C equals?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but as from just now eqn 4 = A(x^2+4) + x(Bx+c) can i sub X = 0 for Bx+c(x) so that 4=A(0^2+4)? which turn out to be 4 = 4A?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Bx^2+Cx...that is the idea

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ax^2+4A+Bx^2+Cx, ok we should group these with like terms > (A+B)*x^2 +(C)*x + 4A

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(A+B)*x^2 +(C)*x + 4A = 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you figured out that A = 1 by looking at the case when x=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and that where i struck finding b and c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now, on the right side here there are no terms with x^2 nor x...so what do you think C would be equal to?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now, you know A=1, C=0, now...to find B....so we look at where we have B, only ing (A+B)*x^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which (1 + B) *x^2 because A=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0again, there are no x^2 terms on the other side..so B must equal what?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the idea is that on the other side the term is 4 + 0*x^2, so we have (1+B)*x^2 = (0)*x^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, so what are A, B and C

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one you have them all, substitute them back in the fractions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where they were separted

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in here: 4/(x*(x^2+4))=A/(x)+B/(x^2+4)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok got it.. thx a lot man~ sry to wate such a long time.. hope u don't mind

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no problem, now do you think you can solve another partial fraction problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not really.. cos' theres a lot of type of partial fraction..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well the premise is the same....separate the denominator into a product of functions...then create the letter fractions with these smaller functions...solve for the letters...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just use this conversation as notes....try another and good luck!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.