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Find constants \(A\) and \(B\) such that
\(\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\)
for all \(x\) such that \(x \neq -1\) and \(x\neq2\). Give your answer as the ordered pair \((A,B)\).

 

 

smileysmileysmiley

 Apr 16, 2020
 #1
avatar+24995 
+3

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

 

\(\text{Let $x^2 - x - 2 = (x-2)(x+1)$}\)

 

\(\begin{array}{|rcll|} \hline \dfrac{x + 7}{x^2 - x - 2} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad & | \quad \cdot (x-2)(x+1) \\\\ x + 7 &=& \dfrac{A(x-2)(x+1)}{x - 2} + \dfrac{B(x-2)(x+1)}{x + 1} \quad & | \quad x\ne2,\ x\ne -1 \\\\ x + 7 &=& A (x+1) + B(x-2) \\ \\ \hline x=-1: \quad -1 + 7 &=& A (-1+1) + B(-1-2) \\ \quad 6 &=& 0 -3B \\ \quad 6 &=& -3B \\ \quad -3B &=& 6 \quad | \quad :(-3) \\ \quad B &=& \frac{6}{-3} \\ \quad \mathbf{B} & \mathbf{=} & \mathbf{-2} \\ \hline x=2: \quad 2 + 7 &=& A (2+1) + B(2-2) \\ \quad 9 &=& 3A +0 \\ \quad 9 &=& 3A \\ \quad 3A &=& 9 \quad | \quad :3 \\ \quad A &=& \frac{9}{3} \\ \quad \mathbf{A} & \mathbf{=} & \mathbf{3} \\ \hline \end{array}\)

 

\(\mathbf{(A,\ B) = (3,\ -2)}\)

 

laugh

 Apr 16, 2020
 #2
avatar+20 
+1

Can you go into more detail on why you specifically used the numbers it labeled as undefined?

Joshy1028  Apr 16, 2020
 #3
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+1

Explanation: 

It doesn't depend on the numbers labelled undefined 

but it was the numbers that make one variable to be zero

For example in this question:

\(A(x+1)+B(x-2)\) If x=-1 then 0A=0 so we cancelled one variable 

and then x=2 to cancel B and find A 

Guest Apr 16, 2020
 #4
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+1

https://www.youtube.com/watch?v=BvGKVn-85jM Partial fraction's video, hope it helps! 

Guest Apr 16, 2020

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