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Jadedry
 one year ago
Resolve the following expression into its simplest partial fractions:
Jadedry
 one year ago
Resolve the following expression into its simplest partial fractions:

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Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1the equation in question is: \[(x2) / ( (x^2 +1)*(x1)^2)\] I've managed to figure out that \[0.5 / (x1) ^2 \] is one of the partial fractions, but I'm having a lot of trouble with the others. Thanks in advance!

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1In this particular scenario you can decompose this to following partial fractions, \[\frac{x2}{(x^2+1)(x1)^2} = \frac{Ax+B}{x^2+1}+\frac{C}{(x1)}+\frac{D}{(x1)^2}\] So what you have to do now is to find the values of the A,B,C,D

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0@BAdhi can you please explain why its Ax + B and not simply A or B alone...

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1@Firecat97 Because \[(x^{2} + 1)\] is quadratic due to the fact that x is squared in the case of \[(x1)^{2}\] , the x is not quadratic, but you have to come up with a numerator for \[(x1) \] as well as \[(x1)^{2}\]

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0ohhhhh okayyy thanks! we just started learning this stuff last lesson :)

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1@BAdhi I managed to get that far in my calculations, but I never come up with the right numerators? I think my method may be too complex, is there a simple way to find the numerator when a simple substitute for x will not cancel out all but one constant?

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0okay thanks @Jadedry

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1There are mainly two ways to find the constant First one is the substituting values to the x and comparing numerator Second one is comparing the coefficients of each order of x The numerator is, \(x2 = (Ax+B)(x^22x+1)+C(x^3x^2+x1)+D(x^2+1)\) now you have to compare the coefficients of some of the orders of x (choose the easiest one so that only one unknown or less number of coefficient is left) For example take \(x^3\) \(0 = A+C \implies C = A\) from constant => \(2 = BC+D \implies B = 1.5+C\) compare another single such occation and youll find the value for C and thus values to others from above equations (i recommend coeff of \(x\))

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1Ah, I see you've expanded the equations I didn't do that. I don't understand how you got: 0= A +C I understand the second one, but I'm sorry to say the first one escapes me?

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1the coefficient of \(x^3\) on the left side is 0 the coefficient of the \(x^3\) on the right side comes only from \((Ax+\cdots)(x^2\cdots) +C(x^3+\cdots)+\cdots = (A+C)x^3 +\cdots\) when equating the left side to right, 0 = (A+C)

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0would the value for C be 3/2?

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1@BAdhi Ah that is clearer, thank you. @FireKat97 No, I don't think those are the answers.

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1@FireKat97 I think I came up with those numbers myself, but the textbook says they are wrong, no idea why.

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1@FireKat97 If we take the numerator of (x1)^2 as Cx+D jst like for (x^2+1) \[\frac{Cx+D}{(x1)^2} = \frac{C(x1)+(D+C)}{(x1)^2}=\frac{C(x1)}{(x1)^2}+\frac{(CD)}{(x1)^2}= \frac{C}{(x1)}+\frac{E}{(x1)^2}\] So you can see it can furthur simplfied to that but you cant do it to x^2+1

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1there is a square at for the denominator of E its not visible :(

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1@FireKat97 Still not right, my book says the answer is 1??

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0soooo we have an E too now?

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0wait I think I see why C is 1 but I maybe wrong...

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0so do we get A 2B + C 1 = 0 then C 2(1.5 + C) + C 1 = 0 C + 3  2C +C  1 = 0 2C = 2 C= 1? idkkkk

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1@FireKat97 actually that look pretty good

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1@FireKat97 where did you get the eq ( A 2B + C 1 = 0 ) from though?

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0I did the same thing that @BAdhi was doing earlier, but I did it to the coefficients for x as he had suggested.

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1@FireKat97 Yes there is a E but also note that D has disappeard :P. so if you want you can replace E with D and you'd end up with the expression that i've used on the original question

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0ummmmmmmm soo the E is really just a D?

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0@Jadedry what does your textbook say the whole answer is btw?

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1@FireKat97 it says the answer is: \[\frac{ 1}{ x1 }  \frac{ 1 }{ 2(x1)^2}  \frac{2x+1}{2(x^2+1)}\]

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1so, D is still 0.5 a= 1 b = 0.5 c = 1 right @BAdhi ?

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1ah i think I have it. So if you substitute 2 for x you get: 2a + b+ 5c +5d if you substitute c for a and 1.5 + c for b you have: 2c + c 1.5 +5c  2.5 = 0 =4c  4 = 0  therefore c = 1 as the equation requires therefore, b= 1.5 +c b = 0.5 a = c a = 1 and d remains 0.5 Yeah! got it!

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0Yup I think I got it too, thanks @Jadedry and @BAdhi :D

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1Thanks @BAdhi , never would have got it without that a = c , you also gave pretty solid advice. Thanks @FireKat97 you managed to point me in the right direction Never would have done it without you guys, this problem has been bugging me for days.

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1Okay, closing the board guys. Thanks again!
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