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RolyPoly
 3 years ago
Resolve into partial fraction
\[\frac{x^2}{1x^2}\]
RolyPoly
 3 years ago
Resolve into partial fraction \[\frac{x^2}{1x^2}\]

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RolyPoly
 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??

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

RolyPoly
 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

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

RolyPoly
 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

RolyPoly
 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 \(x=1\) \(1=2B\) \(B=1/2\) \[\frac{1}{(1x)(1+x)}=\frac{1/2}{(1−x)}\frac {1/2}{(1+x)}\] \[\frac{x^2}{(1x)(1+x)}=\frac{x^2}{2(1−x)}\frac {x^2}{2(1+x)}\]

RolyPoly
 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?

RolyPoly
 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))

RolyPoly
 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

RolyPoly
 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.

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

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