Algebra II Quotients of Rational Expressions

[velox_xox]

Homework Statement

Simplify. (p⁴ - q⁴)/(p + q)² ÷ 1/(p² + q²)

Answer: (p - q)/(p + q)

Homework Equations

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The Attempt at a Solution

Transformed it to a multiplication problem.
(p⁴ - q⁴)/(p + q)² X (p² + q²)/1
Difference of the squares in the numerator of the first expression:
(p² + q²)(p² - q²)/(p + q)² X (p² + q²)/1
Difference of the squares (once more) in the numerator of the first expression:
(p² + q²)(p + q)(p - q)/(p + q)² X (p² + q²)/1
Simplified the (p +q) from the numerator and denominator of the first equation:
(p² + q²)(p -q)/(p + q) X (p² + q²)/1
Simplified the (p² + q²) from the numerators of both expressions:
(p - q)/(p + q)

My question is with the part. Is that proper form? A.k.a. Did I not bend the rules of algebra to get the correct answer? And if so, can someone give example of when simplifying both numerators is okay or not okay?

Thank you!

 

[Dick]

Now if (p^2+q^2) is in the numerators of both expressions how can you just cancel it out? Shouldn't your answer have (p^2+q^2)^2 in it?

 

[velox_xox]

That's what I thought, but the textbook's answer is (p - q)/(p +q), so either it's incorrect; Or somewhere I solved it wrong in the first place, and now I'm just bending the rules. That's why I'm asking. :)

 

[Dick]

That's what I thought, but the textbook's answer is (p - q)/(p +q), so either it's incorrect; Or somewhere I solved it wrong in the first place, and now I'm just bending the rules. That's why I'm asking. :)

Well, I get (p-q)*(p^2+q^2)^2/(p+q). I don't think you are doing anything wrong except for the bad cancellation to match the books answer. Either the books answer is wrong or the problem is misstated.

 

[velox_xox]

Nope, I've checked it like seven times now to be sure (both the problem and the answer). The only thing that is different in my writing of the problem from the actual textbook's version is that it is actually a fraction within a fraction.

(p⁴ - q⁴)/(p + q)² <<This part being numerator
(p² + q²) <<This part being denominator

That shouldn't make a difference, though. Right?

...Anyone else want to give it a go?

 

[scurty]

Nope, I've checked it like seven times now to be sure (both the problem and the answer). The only thing that is different in my writing of the problem from the actual textbook's version is that it is actually a fraction within a fraction.

(p⁴ - q⁴)/(p + q)² <<This part being numerator
(p² + q²) <<This part being denominator

That shouldn't make a difference, though. Right?

...Anyone else want to give it a go?

Let's try writing them as fractions. From the post I quoted I get the impression the fraction is: \displaystyle\frac{\frac{p^4-q^4}{(p+q)^2}}{p^2+q^2}. If this is the case, you gave us the wrong problem to work with in your original post!

\displaystyle\frac{\frac{p^4-q^4}{(p+q)^2}}{p^2+q^2} = \frac{p^4-q^4}{(p+q)^2} \cdot \frac{1}{p^2+q^2} \neq \frac{p^4-q^4}{(p+q)^2} \div \frac{1}{p^2+q^2}

 

[velox_xox]

*facepalm* Oh, that is so obvious! Like how 8 and 1/8 aren't the same.

I was able to get the correct answer with that. Also, for future reference, how am I supposed to write a fraction in the numerator of a fraction?? Like what you did? (The reason I didn't write that it that way in the first place is because I didn't know how to do it.)

Thanks scurty for your insight, and sorry Dick for the mistake!

 

[Dick]

*facepalm* Oh, that is so obvious! Like how 8 and 1/8 aren't the same.

I was able to get the correct answer with that. Also, for future reference, how am I supposed to write a fraction in the numerator of a fraction?? Like what you did? (The reason I didn't write that it that way in the first place is because I didn't know how to do it.)

Thanks scurty for your insight, and sorry Dick for the mistake!

No problem. Just use an extra set of parentheses. Like ((p^4-q^4)/(p+q)^2))/(p^2+q^2). Or learn to Latex it. That's always nice.

 

[velox_xox]

Got it. I'll be sure to do that in the future. As for Latex, I have no idea how to do that. Is there a section on how to learn on the forum? :D

And a very much belated but definitely deserved thanks to scurty and Dick. Thank you!

 

[scurty]

No problem!

Check here for a LaTeX guide: https://www.physicsforums.com/showthread.php?t=546968