Solving for Q and R in Seemingly Unsolvable Problem

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Homework Help Overview

The discussion revolves around a mathematical problem involving the decomposition of a rational expression into partial fractions. The original poster describes a scenario where they need to solve for two unknowns, Q and R, given a specific equation that appears to have more unknowns than equations.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore the simplification of the equation and question the nature of the unknowns Q and R. There is a suggestion that the problem may relate to partial fractions, and some participants express uncertainty about the setup and the number of variables involved.

Discussion Status

Some participants have offered guidance regarding the concept of partial fractions, which seems to have helped the original poster gain clarity. However, there is still a lack of consensus on the interpretation of the problem and the nature of the unknowns.

Contextual Notes

The original poster notes that the problem is not part of their homework but arises from a test for someone else, which may influence their approach and understanding of the context. There is also mention of potential confusion regarding the terminology used in engineering versus physics.

erok81
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This isn't from my homework, but it came up during a test for my sister's husband. He works at an Air Force base as some type of engineer and they are required to take this general math class. So as far as I can tell, it's nothing advanced but I can't figure it out, and swear it is unsolvable - three unknowns and only one equation.

Maybe it's something I don't know. The professor gave an answer for both Q and R, nothing for x.

Homework Statement



Solve for Q and R.

[itex]\frac{(x+4)}{(x+1)}(x+2) = \frac{Q}{(x+1)} + \frac{R}{(x+2)}[/itex]

Homework Equations



Nothing.

The Attempt at a Solution



So I got it simplified down to x+4 = Q(x+2) + R(x+1) but that's about as far as I can get. The only thing I can think of is Q and R are some engineering terms I've never heard of (physics student here). I tried solving it for x, but that didn't get me anywhere

Q and R end up being regular old whole numbers. I can post those values as well.
 
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erok81 said:
This isn't from my homework, but it came up during a test for my sister's husband. He works at an Air Force base as some type of engineer and they are required to take this general math class. So as far as I can tell, it's nothing advanced but I can't figure it out, and swear it is unsolvable - three unknowns and only one equation.

Maybe it's something I don't know. The professor gave an answer for both Q and R, nothing for x.

Homework Statement



Solve for Q and R.

[itex]\frac{(x+4)}{(x+1)}(x+2) = \frac{Q}{(x+1)} + \frac{R}{(x+2)}[/itex]

Homework Equations



Nothing.

The Attempt at a Solution



So I got it simplified down to x+4 = Q(x+2) + R(x+1) but that's about as far as I can get. The only thing I can think of is Q and R are some engineering terms I've never heard of (physics student here). I tried solving it for x, but that didn't get me anywhere

Q and R end up being regular old whole numbers. I can post those values as well.

It looks like it's just a partial fractions question. You know those, don't you?
 
I am assuming this is the equation you have:

[tex] <br /> \frac{x+4}{(x+1)(x+2)} = \frac{P}{x+1} + \frac{Q}{x+2}<br /> [/tex]

This type of decomposition is useful when integrating the quotient of two polynomial functions.
It is typically called "partial fractions decomposition".

Here's an example:

Suppose you had

[tex] \frac{5x-4}{(x+1)(2x-1)} = \frac{P}{x+1} + \frac{Q}{2x-1}[/tex]

You can multiply both sides by

[tex] (x+1)(2x-1)[/tex]

to get

[tex] 5x-4 = P(x+1) + Q(2x-1)[/tex]

Now, you can expand the above, collect like-terms, and then compare coefficients of both sides:

[tex] 5x-4 = (P+2Q)x + (P-Q)[/tex]

We now have a system of 2 equations in 2 unknowns:

[tex] 5 = P+2Q[/tex]

[tex] -4 = P-Q[/tex]

Does this help?
 
Yes! Thank you both. That helps. I can't believe I didn't even think of trying partial fractions. I just saw so many variables and had a negative outlook on it from the inset. Plus I never saw the original problem, so I also assumed something was left off.
 

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