am I on the right track here?

[allegansveritatem]

Here is problem:
I am sorry the image is blurred a bit on top but I think the sense of the problem can be made out.
Here is how I tried to solve this: Part a):

I wanted a factor in the form (x-1) because I was going too used synthetic division on the result.
I set the above formula equal to the volume of the tablet which is .25pi/3. So my plan is to work out the equation and use synthetic division on the result. My problem is that I have already worked this out about 4 times and have not got any realistic results. When I use synthetic division with my factor ie, (r-4/9) I keep getting a remainder.
So...what is my problem here? I must have worked on this for 2.5 hours this afternoon but could not get things to shake out.

 

[Dr.Peterson]

The last line makes no sense at all, though everything before that is fine.

What is x??? You don't need it, and you haven't defined it.

The next to last line is the answer to part (a), depending on what form you like. For part (b), just set your formula from part (a) equal to the desired volume, pi/12.

Don't try to solve the equation until you've written it out, in the form I just stated. It will be a cubic; it isn't worth factoring until you have the entire equation in standard form. I would first multiply everything by 12 to clear fractions, and maybe then try synthetic division. Don't rush it.

 

[allegansveritatem]

The last line makes no sense at all, though everything before that is fine.

What is x??? You don't need it, and you haven't defined it.

The next to last line is the answer to part (a), depending on what form you like. For part (b), just set your formula from part (a) equal to the desired volume, pi/12.

Don't try to solve the equation until you've written it out, in the form I just stated. It will be a cubic; it isn't worth factoring until you have the entire equation in standard form. I would first multiply everything by 12 to clear fractions, and maybe then try synthetic division. Don't rush it.

thank you for looking it over. The x in the last line is actually a multiplication sign...I was using the sign just to keep it clear to myself what I was trying to do. As for the term -4/6(r-9/4), I did that, as I say, because I want to have a factor I can use to do synthetic division with--one reason being that this problem comes at the end of a section dealing with synthetic division. But I am not sure my reasoning is correct here. Am I doing something crazy? I have been influenced in this pursuit of the right factor by reading in the solutions manual where my author says upon occasion, after obtaining a fancy looking cubic polynomial, that since such and such an (x-1) or (x+3) etc., is a factor can use synthetic division. This is one of the points I would like to ask about, namely, when to use synthetic division and when to try something else.

 

[allegansveritatem]

Something else: I noticed that when I squared r, which is .5 cm, I got .25 cm, so that I am getting a figure that is smaller than its square root! What? I am thinking I should convert the units to mm or I am going to come out with something nutty. No?

 

[Steven G]

If 0<a<1, then YES sqrt(a^2) < sqrt(a). If you used fractions! instead of decimals you would see this. For example sqrt(1/4) = 1/2 and 1/2> 1/4 !!!!!

Yes, when it comes to finding roots it is very good to have a common factor. But if you think about it (r-9/4) might not be a factor for the other volume.

For example IF the other volume was r^2 + 5r+ 6, then you will have r^2 + 5r+ 6 = -pi r^2*4/6*(r-9/4). The thing is when you bring everything to one side there is NO guarantee that (r-9/4) will be a factor. You really need to think this one through so you do not do the factoring too early as you did above.

 

[Dr.Peterson]

Something else: I noticed that when I squared r, which is .5 cm, I got .25 cm, so that I am getting a figure that is smaller than its square root! What? I am thinking I should convert the units to mm or I am going to come out with something nutty. No?

Actually, you didn't get .25 cm; you got .25 cm^2. You can't really compare a length with an area.

In terms of mere fractions, as has been said, the square of 1/2 is 1/4, and that's just the way things work.

As for the term -4/6(r-9/4), I did that, as I say, because I want to have a factor I can use to do synthetic division with--one reason being that this problem comes at the end of a section dealing with synthetic division. But I am not sure my reasoning is correct here. Am I doing something crazy? I have been influenced in this pursuit of the right factor by reading in the solutions manual where my author says upon occasion, after obtaining a fancy looking cubic polynomial, that since such and such an (x-1) or (x+3) etc., is a factor can use synthetic division. This is one of the points I would like to ask about, namely, when to use synthetic division and when to try something else.

Now, go back to the equation, [MATH]\pi r^2 \frac{9 - 4r}{6} = \frac{\pi}{12}[/MATH], and start solving it, without trying to factor too soon. Multiply both sides by 12 to clear fractions, divide by pi, get a zero on one side, and then see if you can find a potential factor to divide by.

The place to use synthetic division is when you have reason to think you have a factor; and one way to decide that is to list potential rational zeros. Don't randomly pick a factor; and don't factor until there's a zero on one side.

Let's see your work.

 

[allegansveritatem]

Something else: I noticed that when I squared r, which is .5 cm, I got .25 cm, so that I and getting a figure that is smaller than it's square root! What? I am thinking I should convert the units to mm or I am going to come out with something nutty. No?

If 0<a<1, then YES sqrt(a^2) < sqrt(a). If you used fractions! instead of decimals you would see this. For example sqrt(1/4) = 1/2 and 1/2> 1/4 !!!!!

Yes, when it comes to finding roots it is very good to have a common factor. But if you think about it (r-9/4) might not be a factor for the other volume.

For example IF the other volume was r^2 + 5r+ 6, then you will have r^2 + 5r+ 6 = -pi r^2*4/6*(r-9/4). The thing is when you bring everything to one side there is NO guarantee that (r-9/4) will be a factor. You really need to think this one through so you do not do the factoring too early as you did above.

Well I will have to think it through, as you say. Thanks

 

[allegansveritatem]

Actually, you didn't get .25 cm; you got .25 cm^2. You can't really compare a length with an area.

In terms of mere fractions, as has been said, the square of 1/2 is 1/4, and that's just the way things work.

Now, go back to the equation, [MATH]\pi r^2 \frac{9 - 4r}{6} = \frac{\pi}{12}[/MATH], and start solving it, without trying to factor too soon. Multiply both sides by 12 to clear fractions, divide by pi, get a zero on one side, and then see if you can find a potential factor to divide by.

The place to use synthetic division is when you have reason to think you have a factor; and one way to decide that is to list potential rational zeros. Don't randomly pick a factor; and don't factor until there's a zero on one side.

Let's see your work.

I think what I was thinking was something like: Well on one side of the = sign I have a set of factors and on the other I have another set of factors and somehow I had conveniently ignored the fact that the intervention of that = sign insulated, so to speak, the...what? the distributive reach of the factors on one side of the sign from those on the other side. In other words, I was treating the two sets of factors as one.
Yes, I will continue on this tomorrow and post my results. Thanks for the tips.

 

[allegansveritatem]

I worked this out today and got this far:


So...I just couldn't seem to figure out how to factor this or get the solution other than use the calculator. I tried to factor this way:


but this doesn't seem to lead to where I want to go. I have a feeling I am forgetting something I already know here...Anyway, how do I solve this without a calculator graph?

 

[Dr.Peterson]

You appear to be thrashing around trying random bits of factoring, forgetting that you have to factor the entire left hand side, or it's worthless.

Recall my suggestion:

The place to use synthetic division is when you have reason to think you have a factor; and one way to decide that is to list potential rational zeros. Don't randomly pick a factor; and don't factor until there's a zero on one side.

You have the right equation, [MATH]8r^3 - 18r^2 + 1 = 0[/MATH]. You had mentioned synthetic division, so I assumed you have learned about the rational root theorem, which helps you find likely factors to divide by. The possible rational roots are 1/2, 1/4, 1/8 and their negatives. Sound familiar? That gives you six possible factors to try. Do so.

 

[allegansveritatem]

You appear to be thrashing around trying random bits of factoring, forgetting that you have to factor the entire left hand side, or it's worthless.

Recall my suggestion:

You have the right equation, [MATH]8r^3 - 18r^2 + 1 = 0[/MATH]. You had mentioned synthetic division, so I assumed you have learned about the rational root theorem, which helps you find likely factors to divide by. The possible rational roots are 1/2, 1/4, 1/8 and their negatives. Sound familiar? That gives you six possible factors to try. Do so.

no, I don't know that one. Let me look it up and see what it is about.
One thing I hit upon after I posted today while thinking about this problem: I could just estimate what r probably is and plug some probailities into the equation until I hit on the right one(s) but that ain't algebra, is it?
And, yes, I know that it's no good unless all the terms on the side opposite the zero are linked by multiplication signs.

 

[Dr.Peterson]

I need to know what you have learned, in order to suggest what to do. You said, "this problem comes at the end of a section dealing with synthetic division;" what besides synthetic division was covered? What did they do with synthetic division?

 

[allegansveritatem]

You appear to be thrashing around trying random bits of factoring, forgetting that you have to factor the entire left hand side, or it's worthless.

Recall my suggestion:

You have the right equation, [MATH]8r^3 - 18r^2 + 1 = 0[/MATH]. You had mentioned synthetic division, so I assumed you have learned about the rational root theorem, which helps you find likely factors to divide by. The possible rational roots are 1/2, 1/4, 1/8 and their negatives. Sound familiar? That gives you six possible factors to try. Do so.

I just watched a video on the rational root theorem. That looks easy enough, especially with the polynomial that is in question here. I will give it a shot.

 

[allegansveritatem]

I need to know what you have learned, in order to suggest what to do. You said, "this problem comes at the end of a section dealing with synthetic division;" what besides synthetic division was covered? What did they do with synthetic division?

The theorems covered were 1) Remainder Theorem and one of its corollaries or special cases, 2) the Factor Theorem.

 

[Dr.Peterson]

Interesting. The Rational Root Theorem is commonly taught (and used) along with those; it provides possible roots to test by the Factor Theorem. I can't think of any other way to solve this cubic equation that you would be expected to use, apart from just graphing it, or just guessing.

 

[allegansveritatem]

Interesting. The Rational Root Theorem is commonly taught (and used) along with those; it provides possible roots to test by the Factor Theorem. I can't think of any other way to solve this cubic equation that you would be expected to use, apart from just graphing it, or just guessing.

Well, it wouldn't be hard to guess this one--you could set up a ratio, something like: 1.5 is to .333 as .5 is to x. and you would get a ballpark figure to start with.
No, that will not do. I just tried it and got something about half as large as needed.

 

[Otis]

… The x in the last line is actually a multiplication sign …

Hi. We don't generally use × as a multiplication symbol, when writing algebra. (Scientific notation is one exception.)

? You can write grouping symbols or a centered dot.

Pi r² (1/6)

Pi∙r²∙1/6

\(\;\)

 

[allegansveritatem]

but, if I use the ratio .5/0.3333=x/1.5 I get 2.25 and this IS a good approximation of the right answer. .................edited

 

[allegansveritatem]

No that's not right either..unless the units are changed to mm. I see that a ratio won't work.Unless it is some kind of inverse ratio--i

 

[Deleted member 4993]

but, if I use the ratio .5/3333=x/1.5 I get 2.25 and this IS a good approximation of the right answer.

The function (derivative) in question does NOT have a rational root at 2.25. By numerical approximation (Newton's method) we get x ~ 2.224744873 ~ 20/9 (not exact) as one of the roots......... edited