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Find constants A and B such that

 

(x+7)/(x²-x-2)+a/(x-2)+b/(x+1)


for all x such that x is not equal to -1 and x is not equal to 2. Give your answer as the ordered pair (A,B).

 Aug 14, 2020
 #1
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+1

Notice that x-2 and x+1 are factors of x^2-x-2.

 

\(\frac{x+7}{x^2-x-2}=\frac{a}{x-2}+\frac{b}{x+1}\\ \frac{x+7}{x^2-x-2}=\frac{a(x+1)}{x^2-x-2}+\frac{b(x-2)}{x^2-x-2}\\ x-7=a(x+1)+b(x-2)\\ x-7=ax+a+bx-2b\\\)

ax+bx must add up to the x term on the left side, which is x. a-2b must subtract to the constant on the left side, -7. We then have a system of equations.

 

a+b=1

a-2b=-7

 

We find that a= -5/3 and b=8/3.

 Aug 14, 2020
 #2
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Sorry, wrong answer!!

Guest Aug 14, 2020
 #3
avatar+110762 
+1

\(\frac{(x+7)}{(x²-x-2)}+\frac{a}{x-2}+\frac{b}{(x+1)}\)

 

Such that what?  Is there meant to be an equal sign somewhere?

 Aug 15, 2020
 #4
avatar+30931 
+2

I assume this is a partial fraction question as interpreted by Guest#1.  Guest#1 has the right approach, but unfortunately mistakenly set a - 2b = -7, where it should be a - 2b = 7.  Given this correction you shoud be able to determine a and b.

 Aug 15, 2020

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