happykiddo
  • happykiddo
Partial Fractions. Could someone please explain how I could use elimination to find the values A,B,C,D? Attached problem and solution...
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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happykiddo
  • happykiddo
1 Attachment
happykiddo
  • happykiddo
I don't like the substitution method because it seems like they just guess the values for A & B.
freckles
  • freckles
you mean the Heaviside method?

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freckles
  • freckles
They randomnly chosen x values to solve for A,B,C, and D. say you have 5+2x=A(x+1)+B(x-1) You can solve this for A and B...by choosing values of x such that the equation is easier to solve for A and B like choosing x=1 and x=-1 like since 5+2x=A(x+1)+B(x-1) is an equality it must hold for all x values... So choosing x=-1 and x=1 is fine to solve for A and B. input x=-1: 5+2(-1)=A(-1+1)+B(-1-1) 5-2=A(0)+B(-2) 3=B(-2) B=-3/2 And input x=1: 5+2x=A(x+1)-3/2(x-1) 5+2=A(1+1)-3/2(1-1) 7=A(2) A=7/2 so we have 5+2x=7/2(x+1)-3/2(x-1) you can expand the right hand side to test to see if this is indeed true 7/2(x+1)-3/2(x-1) = 7/2 x+7/2-3/2 x +3/2 4/2x+10/2 2x+5 and yes 2x+5 is the same as 5+2x
freckles
  • freckles
however if you don't like Heaviside method you could just solve a system of linear equations but sometimes you have to do this even with Heaviside method.
happykiddo
  • happykiddo
okay, so looking at the problem I gave how did they choose the values 1/4 & -1/4 for A and B? Why not 1 and -1 that would also equal 0...
freckles
  • freckles
\[A(x+1)(x-1)^2+B(x-1)(x+1)^2+C(x-1)^2+D(x+1)=1 \\ A(x^2-1)(x-1)+B(x^2-1)(x+1)+C(x^2-2x+1)+D(x+1)=1 \\ A(x^3-x^2-x+1)+B(x^3+x^2-x-1)+C(x^2-2x+1)+D(x+1)=1 \\ x^3(A+B)+x^2(-A+B+C)+x(-A-B-2C+D)+(A-B+C+D)=1 \\ \text{ so you have the follow system \to solve } \\ A+B=0 \\ -A+B+C=0 \\ -A-B-2C+D=0 \\ A-B+C+D=1\]
freckles
  • freckles
so you want to do heaviside method?
happykiddo
  • happykiddo
If it will make more sense, lets do it.
freckles
  • freckles
I don't know which makes more sense to you...They both make sense to me. \[A(x+1)(x-1)^2+B(x-1)(x+1)^2+C(x-1)^2+D(x+1)^2=1\] notice some of these would be 0 if you choose x=1 or x=-1 So choosing input x=1: do you see how we get D(1+1)^2=1?
freckles
  • freckles
if you don't see how, please let me know
happykiddo
  • happykiddo
Gotcha, I see what you did there, so the input values don't matter only the output. Like "D" would equal 1/4. And I would continue putting in values for x until I have only one variable left and solve for it. Big thank you!!
freckles
  • freckles
yeah starting with x=1 and x=-1 makes things easier since most terms will be 0 So the solving for A and B part are a little more difficult but yep you can choose 2 more x values to get a system of 2 linear equations. \[A(x+1)(x-1)^2+B(x-1)(x+1)^2+\frac{1}{4}(x-1)^2+\frac{1}{4}(x+1)^2=1 \\ \text{ so choose } x=2 \\ A(3)(1)^2+B(1)(3)^2+\frac{1}{4}(1)^2+\frac{1}{4}(2)^2=1\] you will choose another x value and you will get another equation in terms of A and B And then solve the system .
freckles
  • freckles
now that isn't exactly what they did
freckles
  • freckles
it looks like they expanded everything and compared both sides of the equations
freckles
  • freckles
but I think the way I'm suggesting might be less work
happykiddo
  • happykiddo
Great thanks for clarifying.
freckles
  • freckles
ok you try to see if you can get their A and B if you cannot holler at me
happykiddo
  • happykiddo
Going back to the post where you said make x=2, leftover I would have A and B with constants. So how would I isolate each of them to find individual values?
freckles
  • freckles
you need to choose another x value to plug in so you have two linear equations involving 2 different constant variables
freckles
  • freckles
like choose x=-2
freckles
  • freckles
or any x that we haven't already used
freckles
  • freckles
\[3A+9B+\frac{1}{4}+\frac{1}{4}(9)=1 \\ -9A-3B+\frac{1}{4}(9)+\frac{1}{4}=1 \\ \text{ simplifying both equations } \\ 3A+9B=1-\frac{10}{4} =\frac{-6}{4}=\frac{-3}{2} \\ -9A-3B=1-\frac{10}{4}=\frac{-3}{2}\] so we have the following system to solve: \[3A+9B=\frac{-3}{2} \\ -9A-3B=\frac{-3}{2} \\ \text{ we can solve by elimination } \\ \text{ multiply first equation by 3 } \\ 9A+27B=\frac{-9}{2} \\ -9A-3B=\frac{-3}{2} \\ \text{ now add equations } \\ 24B=\frac{-12}{2} \\ 24B=-6 \\ B=\frac{-6}{24} =\frac{-1}{4} \\ \text{ now you can plug into either of the two equations here to find } A\]
happykiddo
  • happykiddo
Oh yeah, I feel stupid for forgetting middle school maths. Thanks I appreciate it.
freckles
  • freckles
np
freckles
  • freckles
And you aren't stupid. :p

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