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Find constants A and B such that (x + 7)/(x^2 - x - 2) = A/(x - 2) + B/(x + 1) for all x such that x ≠ 1 and x ≠ 2. Give your answer as the ordered pair (A,B).

 

If you guys are able to solve this quickly then that would be great!

Guest Sep 23, 2018
 #1
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\(\text{This is known as finding the partial fraction expansion}\\ \dfrac{x+7}{x^2-x-2} = \dfrac{x+y}{(x-2)(x+1)} = \dfrac{A}{x-2}+\dfrac{B}{x-1}\)

 

\(\dfrac{x+7}{(x-2)(x+1)} = \dfrac{A(x+1)+B(x-2)}{(x-2)(x+1)}\\ \forall x : x \neq 2,~-1 \\ x+7 = A(x+1) + B(x-2) \\ x = (A+B)x \Rightarrow \\ 1 = A+B\\ 7 = A -2B \)

 

\(B = 1-A\\ 7 = A - 2(1-A) = 3A-2\\ 3A = 9\\ A=3\\ B = 1-3 = -2\\ \dfrac{x+7}{(x-2)(x+1)}=\dfrac{3}{x-2}-\dfrac{2}{x+1}\)

Rom  Sep 23, 2018
edited by Rom  Sep 23, 2018

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