# Does this look right at all?

#### allegansveritatem

##### Full Member
I am asked to simplify this:

Here is what I got...and this is only the latest in a series:
'

I am full of doubt about this because...why would they have that (4-x^2) in both numerator and denomenator if it wasn't meant to disappear somehow? Anyway, this is my best shot so far.

#### tkhunny

##### Moderator
Staff member
Good effort. Don't be afraid to write things over, rather than messily crossing things out.

When you multiplied by the (6x+1)^(2/3), you did that only in the numerator. Whoops. Multiply by that factor in the denominator, too. You should have been suspicious about this when the fractional exponents just vanished. That IS too good to be true.

Note: When you are in calculus, this expression will make a lot more sense. It's not just a bunch of symbols thrown into a fan.

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#### allegansveritatem

##### Full Member
Right...I should have seen that. You can't operate on the numerator like that without doing likewise underneath and still have the ratio stay the same. I think what happened is when I saw this expression it was so mean and nasty that it shook most of the sense out of my head--not that there was ever much in there to begin with. Thanks for the tip. I will work this out again tomorrow and post the result.

#### allegansveritatem

##### Full Member
Does this one look more like it?

#### allegansveritatem

##### Full Member
I just tested this on my calculator by assigning a number to the variable--it is not correct. Off by 3 tenths. I will go at it again tomorrow.

#### Dr.Peterson

##### Elite Member
Where did the (2) come from on the second line?

#### allegansveritatem

##### Full Member
Where did the (2) come from on the second line?
I made a mistake in copying the original problem. If you look at my original post you will see it. In the problem. as it should have been transcribed, there was a 1/3 and a 6 as factors. In my latest post I overlooked the 6 but the 2 in the rest of the work "includes" the 6, so the work is correct. At the time I posted I did not know it was correct because I was using my solutions manual to check things and the solutions manual only gives solutions for odd numbered problems. But today, after again working this out and getting same answer, I suddenly recalled that my text gives answers for all the problems for review sections (which is where this problem comes from) so I checked it in the book and the book confirmed my result. And what is strange is when I try both the original expression and the simplified expression in my calculator (after I have assigned x a value) I get two different answers. the difference is fairly slight but significant, I would say. How to explain this?

#### JeffM

##### Elite Member
I am asked to simplify this:
View attachment 12199

Here is what I got...and this is only the latest in a series:
'View attachment 12200

I am full of doubt about this because...why would they have that (4-x^2) in both numerator and denominator if it wasn't meant to disappear somehow? Anyway, this is my best shot so far.
Yes, it does look nasty, but tkhunny is correct that this sort of ugliness arises in calculus. Technically, the expression above is the derivative of

$$\displaystyle \dfrac{(6x + 1)^{1/3}}{4 - x^2}.$$

The derivative of a function is the function's "slope."

What I would do initially is to break the fraction into two expressions that will be a bit easier to work with, then simplify, then get common denominators, simplify again, and add. Trying to work on that mess in one fell swoop will drive you bonkers quickly.

$$\displaystyle \dfrac{(4 - x^2)\left( \frac{1}{3} \right )(6x + 1)^{-2/3}(6) - (6x + 1)^{1/3}(-\ 2x)}{(4 - x^2)^2} =$$

$$\displaystyle \dfrac{(4 - x^2)\left( \frac{1}{3} \right )(6x + 1)^{-2/3}(6)}{(4 - x^2)^2} - \dfrac{(6x + 1)^{1/3}(-\ 2x)}{(4 - x^2)^2} =$$

$$\displaystyle \dfrac{2(4 - x^2)}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{(6x + 1)^{1/3}(-\ 2x)}{(4 - x^2)^2} =$$

$$\displaystyle \dfrac{2(4 - x^2)}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{(6x + 1)^{2/3} * (6x + 1)^{1/3}(-\ 2x)}{(6x + 1)^{2/3}(4 - x^2)^2} =$$

$$\displaystyle \dfrac{8 - 2x^2}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{(6x + 1)(-\ 2x)}{(6x + 1)^{2/3}(4 - x^2)^2} =$$

$$\displaystyle \dfrac{8 - 2x^2}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{-\ 12x^2 - 2x}{(6x + 1)^{2/3}(4 - x^2)^2} =$$

$$\displaystyle \dfrac{8 - 2x^2 + 12x^2 + 2x}{(6x + 1)^{2/3}(4 - x^2)^2} = \dfrac{10x^2 + 2x + 8}{(6x + 1)^{2/3}(4 - x^2)^2}$$

I shall test it using x = 0.

$$\displaystyle \dfrac{(4 - 0^2)\left( \frac{1}{3} \right )(6 * 0 + 1)^{-2/3}(6) - (6 * 0 + 1)^{1/3}(-\ 2 * 0)}{1^{2/3} * (4 - 0^2)^2} = \dfrac{2(4)(1)^{-2/3}}{1 * 4^2} = \dfrac{1}{2}.$$

$$\displaystyle \dfrac{10 * 0^2 + 2 * 0 + 8}{(6 * 0 + 1)^{2/3}(4 - 0^2)^2} = \dfrac{8}{1 * 16} = \dfrac{1}{2}.$$

Your algebra looks good to me. What was your test number on the calculator?

#### allegansveritatem

##### Full Member
Yes, it does look nasty, but tkhunny is correct that this sort of ugliness arises in calculus. Technically, the expression above is the derivative of

$$\displaystyle \dfrac{(6x + 1)^{1/3}}{4 - x^2}.$$

The derivative of a function is the function's "slope."

What I would do initially is to break the fraction into two expressions that will be a bit easier to work with, then simplify, then get common denominators, simplify again, and add. Trying to work on that mess in one fell swoop will drive you bonkers quickly.

$$\displaystyle \dfrac{(4 - x^2)\left( \frac{1}{3} \right )(6x + 1)^{-2/3}(6) - (6x + 1)^{1/3}(-\ 2x)}{(4 - x^2)^2} =$$

$$\displaystyle \dfrac{(4 - x^2)\left( \frac{1}{3} \right )(6x + 1)^{-2/3}(6)}{(4 - x^2)^2} - \dfrac{(6x + 1)^{1/3}(-\ 2x)}{(4 - x^2)^2} =$$

$$\displaystyle \dfrac{2(4 - x^2)}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{(6x + 1)^{1/3}(-\ 2x)}{(4 - x^2)^2} =$$

$$\displaystyle \dfrac{2(4 - x^2)}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{(6x + 1)^{2/3} * (6x + 1)^{1/3}(-\ 2x)}{(6x + 1)^{2/3}(4 - x^2)^2} =$$

$$\displaystyle \dfrac{8 - 2x^2}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{(6x + 1)(-\ 2x)}{(6x + 1)^{2/3}(4 - x^2)^2} =$$

$$\displaystyle \dfrac{8 - 2x^2}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{-\ 12x^2 - 2x}{(6x + 1)^{2/3}(4 - x^2)^2} =$$

$$\displaystyle \dfrac{8 - 2x^2 + 12x^2 + 2x}{(6x + 1)^{2/3}(4 - x^2)^2} = \dfrac{10x^2 + 2x + 8}{(6x + 1)^{2/3}(4 - x^2)^2}$$

I shall test it using x = 0.

$$\displaystyle \dfrac{(4 - 0^2)\left( \frac{1}{3} \right )(6 * 0 + 1)^{-2/3}(6) - (6 * 0 + 1)^{1/3}(-\ 2 * 0)}{1^{2/3} * (4 - 0^2)^2} = \dfrac{2(4)(1)^{-2/3}}{1 * 4^2} = \dfrac{1}{2}.$$

$$\displaystyle \dfrac{10 * 0^2 + 2 * 0 + 8}{(6 * 0 + 1)^{2/3}(4 - 0^2)^2} = \dfrac{8}{1 * 16} = \dfrac{1}{2}.$$

Your algebra looks good to me. What was your test number on the calculator?
well, I just tried it again and this time the expressions produced the same result...so I must have made some kind of mistake when I input the data to the calculator. So...a happy ending. Thanks for checking it.