#### allegansveritatem

##### Full Member

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- Thread starter allegansveritatem
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Why did the power of (2x-5) which is -1/2 become 1/2 and now is the power of the 5?? Fix that and you probably will have the correct answer. Please post back with your updated answer.

Please do not use X to denote multiplication as you are using X for the variable.

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\(\displaystyle a^5b^{-\tfrac{1}{2}}(a+b)=a^6b^{-\tfrac{1}{2}}+a^5b^{\tfrac{1}{2}}\)This is part of a problem that I have been breaking my head against for 2 hours today. Please tell me if the following is correct and if not, why not:

View attachment 11451

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You are right. I will supply the whole problem now. I posted the above just after I had gotten done working on this problem and really didn't know what I was doing. I thought the work above meant something but when I look at it now I can't remember what. Here is the whol;e problem worked out in the solutions manual:

Why did the power of (2x-5) which is -1/2 become 1/2 and now is the power of the 5?? Fix that and you probably will have the correct answer. Please post back with your updated answer.

Please do not use X to denote multiplication as you are using X for the variable.

This is supposed to be simplified. I am not sure what technique is being employed here. I have never learned it. I have been looking at videos on Youtube and came upon several that seem to fill the bill. This technique entails finding the greatest common factor...but I think the author is telescoping his presentation and certain steps are left out. No?

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well, my problem centered around not being able to see how he could take two terms from different sides of the plus sign and then use them as multiplier for a binomial on the other side of the plus sign and still get the result he gets. I spent a long time going through this process and tried to understand it by constructing simple models but didn't really come to any satisfactory results. Part of the problem is I have never done this kind of problem before. It was explained in the text but not too extensively. I am going to go over this again today and will shake it like a dog shakes a snake--I was going to say rabbit, but in this case snake fits the situation a little better.The answer is correct and I feel that all steps are included. More importantly what steps do you think are not clear.

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That's called the "distributive law": a(b+ c)= ab+ ac.

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Those steps are an example of factoring. Did you see post #3, where pka showed you a simplified form (substituting symbols a and b for the given binomials)?… I am not sure what technique is being employed here. I have never learned it …

Let us know, if you have issues understanding that post.

PS: In the future, please also provide the given exercise and its instructions.

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yes, I knew the author was factoring but I discovered today that I was misunderstanding a very crucial step--when the one factor was taken (whole) from the right side of the sum I kept missing the point that there was a one left in its place. The one then disappeared because it was a factor of something else that absorbed without leaving a trace. So, I kept puzzling over how he was taking that factor from somewhere and leaving a hole in its place. Once I got that something was being left in its place, these problems suddenly became intelligible to me. When I finally saw this, when the scales fell from my eyes, I raised a silent cry of "Eureka!".Those steps are an example of factoring. Did you see post #3, where pka showed you a simplified form (substituting symbols a and b for the given binomials)?

Let us know, if you have issues understanding that post.

PS: In the future, please also provide the given exercise and its instructions.

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Silent? Next time, run outside naked and scream loudly. That's more fun.… I raised a silent cry of "Eureka!".

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I have some pills that would help you with that.Silent? Next time, run outside naked and scream loudly. That's more fun.

-Dan

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too old for that to be interesting for either myself or anyone else.Silent? Next time, run outside naked and scream loudly. That's more fun.

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good old Arky!

Last edited:

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\(\displaystyle (2x-5)^{-\frac{1}{2}} \ \ne (2x-5^{\frac{1}{2}})\)This is part of a problem that I have been breaking my head against for 2 hours today. Please tell me if the following is correct and if not, why not:

View attachment 11451

Thus the second term of your second line is incorrect.

In addition to that, you may want to pay a bit more attention to your presentation - so that we can decipher it without squinting!!

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I agree. I will try to make things neater. As for what I posted in that photo...I think knew what I wanted to present when I presented it...but only God knows now.\(\displaystyle (2x-5)^{-\frac{1}{2}} \ \ne (2x-5^{\frac{1}{2}})\)

Thus the second term of your second line is incorrect.

In addition to that, you may want to pay a bit more attention to your presentation - so that we can decipher it without squinting!!