How can those two minus x squares disappear like that? Who is right?

A

allegansveritatem

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The following is a snapshot from Robert Blixer's College Algebra of the working out of a subtraction problem:

problem2.jpg

How can those two minus x squares disappear like that? I mean, I think the solution should be this:

(8-2x^2)/(x^2-4x-3)
 
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allegansveritatem said:
The following is a snapshot from Robert Blixer's College Algebra of the working out of a subtraction problem:

View attachment 10278

How can those two minus x squares disappear like that? I mean, I think the solution should be this:

(8-2x^2)/(x^2-4x-3)
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You are of course correct that the text is in error.

5x−x2x2−4x−3−3x−x23+4x−x2=5x−x2x2−4x−3−3x−x23+4x−x2∗− 1− 1=\displaystyle \dfrac{5x - x^2}{x^2 - 4x - 3} - \dfrac{3x - x^2}{3 + 4x - x^2} = \dfrac{5x - x^2}{x^2 - 4x - 3} - \dfrac{3x - x^2}{3 + 4x - x^2} * \dfrac{-\ 1}{-\ 1} =

5x−x2x2−4x−3−x2−3xx2−4x−3=5x−x2−(x2−3x)x2−4x−3=5x−(− 3x)−x2−x2x2−4x−3=\displaystyle \dfrac{5x - x^2}{x^2 - 4x - 3} - \dfrac{x^2 - 3x}{x^2 - 4x - 3} = \dfrac{5x - x^2 - (x^2 - 3x)}{x^2 - 4x - 3} = \dfrac{5x - (-\ 3x) - x^2 - x^2}{x^2 - 4x - 3} =

8x−2x2x2−4x−3=2x(4−x)x2−4x−3.\displaystyle \dfrac{8x - 2x^2}{x^2 - 4x - 3} = \dfrac{2x(4 - x)}{x^2 - 4x - 3}.

So there is a mistake somewhere in the text. There is a reference to returning to the original problem. One possibility is that it did not get transcribed correctly. Of course, there are many possibilities, some of them involving demon rum.
 
JeffM said:
You are of course correct that the text is in error.

5x−x2x2−4x−3−3x−x23+4x−x2=5x−x2x2−4x−3−3x−x23+4x−x2∗− 1− 1=\displaystyle \dfrac{5x - x^2}{x^2 - 4x - 3} - \dfrac{3x - x^2}{3 + 4x - x^2} = \dfrac{5x - x^2}{x^2 - 4x - 3} - \dfrac{3x - x^2}{3 + 4x - x^2} * \dfrac{-\ 1}{-\ 1} =

5x−x2x2−4x−3−x2−3xx2−4x−3=5x−x2−(x2−3x)x2−4x−3=5x−(− 3x)−x2−x2x2−4x−3=\displaystyle \dfrac{5x - x^2}{x^2 - 4x - 3} - \dfrac{x^2 - 3x}{x^2 - 4x - 3} = \dfrac{5x - x^2 - (x^2 - 3x)}{x^2 - 4x - 3} = \dfrac{5x - (-\ 3x) - x^2 - x^2}{x^2 - 4x - 3} =

8x−2x2x2−4x−3=2x(4−x)x2−4x−3.\displaystyle \dfrac{8x - 2x^2}{x^2 - 4x - 3} = \dfrac{2x(4 - x)}{x^2 - 4x - 3}.

So there is a mistake somewhere in the text. There is a reference to returning to the original problem. One possibility is that it did not get transcribed correctly. Of course, there are many possibilities, some of them involving demon rum.
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Yes, the book is wrong. Maybe the author overlooked one of the minus signs. I must say it is hard when you can't trust your book....it's almost like not being able to trust the Supreme Court.
 
allegansveritatem said:
Yes, the book is wrong. Maybe the author overlooked one of the minus signs. I must say it is hard when you can't trust your book....it's almost like not being able to trust the Supreme Court.
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Out of curiosity, would you be able to post the whole problem and solution, so we can see the context and have a better idea what happened?

It is not uncommon for textbooks to have errors in exercises or their solutions, but this appears to be more egregious than that. On the other hand, an author like this one is probably no longer responsible for everything in his books, which have become an industry. I would not be surprised if there are many people at the publisher, or elsewhere, who work at making revisions to the many different versions of a book (yours, we've seen, is quite different from those I've seen in America, so it may be several steps away from the author). In fact, some books continue being revised and published under the same name long after the original author died.

Nevertheless, it may be appropriate to write to the publisher and complain.
 
Dr.Peterson said:
Out of curiosity, would you be able to post the whole problem and solution, so we can see the context and have a better idea what happened?

It is not uncommon for textbooks to have errors in exercises or their solutions, but this appears to be more egregious than that. On the other hand, an author like this one is probably no longer responsible for everything in his books, which have become an industry. I would not be surprised if there are many people at the publisher, or elsewhere, who work at making revisions to the many different versions of a book (yours, we've seen, is quite different from those I've seen in America, so it may be several steps away from the author). In fact, some books continue being revised and published under the same name long after the original author died.

Nevertheless, it may be appropriate to write to the publisher and complain.
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OK. I will get a pic of it tomorrow and post it.
 
Dr.Peterson said:
Yes, it's no more than what has been said: a silly error in the last line or else the first numerator was meant to be 5x + x^2 from the start, and was copied wrong).
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Well, I guess the moral of this story is something like: Don't trust any solution you haven't come by the hard way.
 
allegansveritatem said:
Well, I guess the moral of this story is something like: Don't trust any solution you haven't come by the hard way.
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I'd say the moral is, check your own work step by step, and do the same with others' work. And if you can't check the work, check the result.
 
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