allegansveritatem
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- Jan 10, 2018
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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
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:You really should tell us the instructions which I assume it to distribute. If it to simplify you are way off! You can always plug in an x value of your choice and see if both sides are equal as a check
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.
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.
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 …
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!".
I have some pills that would help you with that.Silent? Next time, run outside naked and scream loudly. That's more fun.
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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!
\(\displaystyle (2x-5)^{-\frac{1}{2}} \ \ne (2x-5^{\frac{1}{2}})\)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
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!!