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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 such x that x ≠ -1, and x ≠ -2. Give your answer as the ordered pair (A,B)

 

 

I've been stuck with this problem, and after a bit of math I've only gotten:

Ax + bx + a -2b = x + 7

Now I'm not sure what to do beyond this point!

Thanks in advance!!

 Mar 30, 2021
 #1
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Using partial fractions, we get that (A,B) = (4,-3).

 Mar 30, 2021
 #2
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That isn't correct sadly :(, I'm not sure I fully understand this problem.

Guest Mar 30, 2021
 #3
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The relationship between the rhs and the lhs is that of an identity.

The rhs has to be identical to the lhs for all values of x.

 

Take then your equation Ax + Bx + A  - 2B = x + 7,

strictly it should be written

\(\displaystyle Ax + Bx + A - 2B \equiv x +7.\)

 

For the two to be identical, (producing identical results for all values of x), it's necessary that the number of x's on the lhs should be the same as the number of x's on the rhs and also that the constants should be the same.

That leads to two equations,

\(\displaystyle A+B=1 \text{ and } A-2B=7, \text{ from which } A=3 \text{ and } B=-2.\)

 

Check then to see that 

\(\displaystyle \frac{3}{x-2}-\frac{2}{x+1}\equiv \frac{x+7}{x^{2}-x-2}.\)

 Mar 30, 2021

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