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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 $x\neq 2$. Give your answer as the ordered pair $(A,B)$.

 Jan 15, 2020
 #1
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0

(A,B) = (5,-4).

 Jan 15, 2020
 #2
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That did not work I think it is wrong

 Jan 15, 2020
 #3
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Help me

 Jan 15, 2020
 #4
avatar+24402 
+2

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 \((A,B)\).

 

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

 

\(\begin{array}{|lrcll|} \hline &\mathbf{ \dfrac{x + 7}{x^2 - x - 2} } &=& \mathbf{ \dfrac{A}{x - 2} + \dfrac{B}{x + 1} } \quad | \quad x^2 - x - 2=(x-2)(x+1) \\\\ &\dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad \times (x-2)(x+1) \\\\ &\dfrac{(x + 7)(x-2)(x+1)}{(x-2)(x+1)} &=& \dfrac{A(x-2)(x+1)}{(x-2)} + \dfrac{B(x-2)(x+1)}{(x+1)} \\\\ & x + 7 &=& A(x+1) + B(x-2) \\\\ \mathbf{x=-1:}& -1 + 7 &=& A(-1+1) + B(-1-2) \\ & 6 &=& A\times 0 -3B \\ & 3B &=& -6 \\ & B &=& -\dfrac{6}{3} \\ & \mathbf{B} &=& -2 \\\\ \mathbf{x=2:}& 2 + 7 &=& A(2+1) + B(2-2) \\ & 9 &=& 3A +B\times 0\\ & 3A &=& 9 \\ & A &=& \dfrac{9}{3} \\ & \mathbf{A} &=& 3 \\ \hline \end{array}\)

 

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

 

check:

\(\begin{array}{|rcll|} \hline \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad A=3,\ B=-2 \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{3}{x - 2} - \dfrac{2}{x + 1} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{3(x+1)-2(x-2)}{(x-2)(x+1)} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{3x+3-2x+4}{(x-2)(x+1)} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{x+7}{(x-2)(x+1)} \ \checkmark \\ \hline \end{array}\)

 

 

laugh

 Jan 15, 2020
 #5
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+1

thanks

 Jan 15, 2020

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