anonymous
  • anonymous
Find A,B,C,a,b,c: below is the equation
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\frac{ 1 }{ x ^{3}-3x ^{2}+2x }=\frac{ A }{ x-a }+\frac{ B }{ x-b }+\frac{C}{ x-c}\]
anonymous
  • anonymous
I tried expanding and simplifying but it was messier than I had thought.
AkashdeepDeb
  • AkashdeepDeb
The hint most probably in this question is that you do not actually have to solve it in the conventional sense. You need to factorize the denominator in the LHS as x(x-1)(x-2) And then because ABCabc are constants you can just relate to the LHS and get the answer. Understanding me? :)

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More answers

anonymous
  • anonymous
So far I've got: [A(x-b)(x-c) + B(x-a)(x-c) +C(x-a)(x-b)]x(x-1)(x-2)=(x-a)(x-b)(x-c). Am I going down the wrong path?
anonymous
  • anonymous
I dont know how to execute it.
anonymous
  • anonymous
so it's likely I have gone down the wrong path.
anonymous
  • anonymous
Please give me another hint!
ganeshie8
  • ganeshie8
\(\large \frac{ 1 }{ x ^{3}-3x ^{2}+2x }=\frac{ A }{ x-a }+\frac{ B }{ x-b }+\frac{C}{ x-c} \)
ganeshie8
  • ganeshie8
\(\large \frac{ 1 }{ x ^{3}-3x ^{2}+2x }=\frac{ A }{ x-a }+\frac{ B }{ x-b }+\frac{C}{ x-c} \) \(\large \frac{ 1 }{ x(x-1)(x-2) }=\frac{ A }{ x-a }+\frac{ B }{ x-b }+\frac{C}{ x-c} \)
anonymous
  • anonymous
OK, will I just randomly asign 0,1,2 to a, b,c? and A+B+C=1 so will I have a system of equations solving for A, B, C?
ganeshie8
  • ganeshie8
exactly !!
ganeshie8
  • ganeshie8
A(x-1)(x-2) + Bx(x-2) + Cx(x-1) = 1 A+B+C = 1 A = 1/2 ....
anonymous
  • anonymous
so is there a missing equation here? I only see 2 equations but am solving for A,B,C.
ganeshie8
  • ganeshie8
\(\large \frac{ 1 }{ x ^{3}-3x ^{2}+2x }=\frac{ A }{ x-a }+\frac{ B }{ x-b }+\frac{C}{ x-c} \) \(\large \frac{ 1 }{ x(x-1)(x-2) }=\frac{ A }{ x-a }+\frac{ B }{ x-b }+\frac{C}{ x-c} \) \(\large \frac{ 1 }{ x(x-1)(x-2) }=\frac{ A(x-1)(x-2) + Bx(x-2)+Cx(x-1) }{ x(x-1)(x-2) } \)
ganeshie8
  • ganeshie8
\(\large \frac{ 1 }{ x ^{3}-3x ^{2}+2x }=\frac{ A }{ x-a }+\frac{ B }{ x-b }+\frac{C}{ x-c} \) \(\large \frac{ 1 }{ x(x-1)(x-2) }=\frac{ A }{ x-a }+\frac{ B }{ x-b }+\frac{C}{ x-c} \) \(\large \frac{ 1 }{ x(x-1)(x-2) }=\frac{ A(x-1)(x-2) + Bx(x-2)+Cx(x-1) }{ x(x-1)(x-2) } \) \(\large 1 = A(x-1)(x-2) + Bx(x-2)+Cx(x-1)\)
ganeshie8
  • ganeshie8
compare x^2, x coefficients and constant terms u should get 3 equations
anonymous
  • anonymous
so one equation is A(x-1)(x-2) +Bx(x-2) + Cx(x-1) =1 and another is A + B + C = 1? I just don't see how to get to A = 1/2
ganeshie8
  • ganeshie8
here is the trick, \(\large 1 = A(x-1)(x-2) + Bx(x-2)+Cx(x-1)\) compare x^2 coefficients both sides wat do u get ?
ganeshie8
  • ganeshie8
left side, x^2 coefficient = 0 right side, x^2 coefficient = ?
ganeshie8
  • ganeshie8
if its not clear, simply expand right side, and write it as polynomial
anonymous
  • anonymous
A(x^2 -3x +2) + B(x^2 -2x) + C(x^2 - x)
ganeshie8
  • ganeshie8
Yes, expand completely
ganeshie8
  • ganeshie8
\(\large 1 = A(x-1)(x-2) + Bx(x-2)+Cx(x-1)\) \(\large 1 = A(x^2-3x+2) + B(x^2-2x)+C(x^2-x)\) \(\large 1 = Ax^2-3Ax+2A + Bx^2-2Bx+Cx^2-Cx\) \(\large 1 = x^2(A+B+C) + x(-3A-2B-C) + 2A\)
ganeshie8
  • ganeshie8
see if that looks right
anonymous
  • anonymous
Yes it does.
ganeshie8
  • ganeshie8
now simply compare x^2, x and constant coefficients both sides
ganeshie8
  • ganeshie8
compare \(x^2\) coefficients 0 = A+B+C ----------------------(1) compoare \(x\) coefficients 0 = -3A-2B-C ----------------(2) compare constant terms 1 = 2A --------------(3)
ganeshie8
  • ganeshie8
3 equations, 3 unknowns, u can solve :) Enjoy..
anonymous
  • anonymous
I see thank you so much.
ganeshie8
  • ganeshie8
np :)

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