A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Resolve into partial fraction
\[\frac{x^2}{1x^2}\]
anonymous
 3 years ago
Resolve into partial fraction \[\frac{x^2}{1x^2}\]

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{x^2}{1x^2} = \frac{x^2}{(1x)(1+x)}\]How to continue??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[x ^{2}/(1x)(1+x)=A/(1x)+B(1+x)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But that doesn't seems to work.. Here's what I mean: A(1+x) + B(1x) = x^2 (AB)x +(A+B) = x^2 => AB =0 , A+B =0 Not reasonable.

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.3The degree of numerator must strictly be less than degree of denominator.

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.3so, one thing u can do is, resolve 1/(1x^2) into partial fractions and then multiply x^2 to both terms.

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.3\(\large \frac{(x^2)}{(1x^2) } = \frac{(1x^2)1}{1x^2} = 1\frac{1}{1x^2}\)

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1A(1+x) + B(1x) = x^2 for x= 1 2 B = x^2 B = x^2/2 for x= 0 A + B = x^2 for x= 1 2A = x^2 A = x^2/2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmmm... 1 1 = 2?! Shouldn't it be \[\frac{(x^2)}{(1x^2) } = \frac{(1x^2)+1}{1x^2} = 1+\frac{1}{1x^2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus A(1+x) + B(1x) = x^2 for x= 1 2 B = (1)^2 B = 1/2 for x= 0 A + B = 0 for x= 1 2A = 1^2 A = 1/2 And it doesn't work here :S

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@hartnn Yes! When can I be as smart as you???!!!!!

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.3with just a little bit of practice :P

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac{x^2}{1x^2}=x^2\times\frac{1}{1x^2} = x^2\times\frac{1}{(1x)(1+x)}\] \[\frac{1}{(1x)(1+x)}=\frac A{(1−x)}+\frac B{(1+x)}=\frac{A(1+x)+B(1x)}{(1x)(1+x)}\] \[1=A(1+x)+B(1x)\] for \(x=1\) \(1=2A\) \(A=1/2\) for \(x=1\) \(1=2B\) \(B=1/2\) \[\frac{1}{(1x)(1+x)}=\frac{1/2}{(1−x)}\frac {1/2}{(1+x)}\] \[\frac{x^2}{(1x)(1+x)}=\frac{x^2}{2(1−x)}\frac {x^2}{2(1+x)}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Excuse.. me.... When x =1 1 = A(1+(1)) + B(1(1)) 1 = 0 + 2B B = 1/2?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=Resolve+into+partial+fraction%3A+x%5E2+%2F+%281x%5E2%29 I don't know how to get that 1/(2(1x))

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Such simple maths is killing me :\ Thanks!!!!!!!!!

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1im not sure about the 1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here it goes~ \[\frac{x^2}{1x^2}\]\[=\frac{(x^2)}{1x^2}\]\[=\frac{(1x^2)+1}{1x^2}\]\[=1+\frac{1}{1x^2}\] \[\frac{1}{1x^2}=\frac{A}{1+x}+\frac{B}{1x}\]A(1x)+B(1+x)=1 A = B = 1/2 So, \[\frac{1}{1x^2}=\frac{1}{2(1+x)}+\frac{1}{2(1x)}\] Thus, \[\frac{x^2}{1x^2}=1+\frac{1}{2(1+x)}+\frac{1}{2(1x)}\]

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1A = B = 1/2 great work @RolyPoly

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.3what u have done is correct.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0:'( Last but not least... \[\frac{x^2}{1x^2}=1+\frac{1}{2(1+x)}\frac{1}{2(x1)}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It needs tons of thousands of millions of practice!! :'( Thanks again!!!
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.