# Is this right?

#### allegansveritatem

##### Full Member
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:

#### Jomo

##### Elite Member
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.

#### pka

##### Elite Member
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
$$\displaystyle a^5b^{-\tfrac{1}{2}}(a+b)=a^6b^{-\tfrac{1}{2}}+a^5b^{\tfrac{1}{2}}$$

#### allegansveritatem

##### Full Member
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.
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:

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?

#### Jomo

##### Elite Member
The answer is correct and I feel that all steps are included. More importantly what steps do you think are not clear.

#### allegansveritatem

##### Full Member
The answer is correct and I feel that all steps are included. More importantly what steps do you think are not clear.
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.

#### HallsofIvy

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

#### Otis

##### Senior Member
… I am not sure what technique is being employed here. I have never learned it …
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.

#### allegansveritatem

##### Full Member
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.

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!".

#### Otis

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

#### topsquark

##### Full Member
Silent? Next time, run outside naked and scream loudly. That's more fun.

-Dan

#### allegansveritatem

##### Full Member
Silent? Next time, run outside naked and scream loudly. That's more fun.

too old for that to be interesting for either myself or anyone else.

good old Arky!

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#### Subhotosh Khan

##### Super Moderator
Staff member
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
$$\displaystyle (2x-5)^{-\frac{1}{2}} \ \ne (2x-5^{\frac{1}{2}})$$

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!!

#### allegansveritatem

##### Full Member
$$\displaystyle (2x-5)^{-\frac{1}{2}} \ \ne (2x-5^{\frac{1}{2}})$$

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!!
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.