Single fraction, lowest terms pt. II

[BlBl]

In an earlier post I intimated that there were more questions to come. As I didn't want to resurrect that old thread with additional questions I figured I'd just start a new one. Anyway continuing where I left off. Here's the problem that immediately followed the other:

\(\displaystyle \dfrac{x^2(x^2+9)^{-1/2}-\sqrt{x^2+9}}{x^2}\)

From the problem before I recognized that I should be focused on making radical expressions into fractional exponents.

\(\displaystyle \dfrac{x^2(x^2+9)^{-1/2}-(x^2+9)^{1/2}}{x^2}\)

After I did that it looked like the numerator could be rewritten in a format like a polynomial with a common factor.

\(\displaystyle \dfrac{x^2(x^2+9)^{-1/2}-(x^2+9)^{1/2}}{x^2}=\dfrac{(x^2+9)^{-1/2}[x^2]}{x^2}\)

That last part is obviously incorrect. I got stuck here figuring out what should go after the common factor. I know that the x squared will be left over, but I'm not sure how to turn the last part into the right format.

EDIT: In writing the problem out and explaining it here, while checking with the book, I figured it out.

\(\displaystyle \dfrac{(x^2+9)^{-1/2}[x^2-(x^2+9)]}{x^2}=\dfrac{(x^2+9)^{-1/2}-9}{x^2}=\dfrac{-9}{x^2(x^2+9)^{1/2}}\)

The next one goes like this:

\(\displaystyle \dfrac{(x^2-9)^{1/3}3-(4x)(\dfrac{1}{3})(x^2-9)^{-2/3}(2x)}{[(x^2-9)^{1/3}]^2}\)

So I start to work on this by looking for the common factor in the numerator. Which I determined to be \(\displaystyle (x^2-9)^{-2/3}\) So I proceeded to write:

\(\displaystyle \dfrac{(x^2-9)^{-2/3}[3-(4x)(\dfrac{1}{3})(2x)]}{[(x^2-9)^{1/3}]^2}\)

I know that that's wrong, but I'm just leaving that as a placeholder. I see from the answer that the book gave that I should have kept the 3 with the common factor, but I need direction here.

EDIT 2: Corrected that first problem to take out the incorrect negative fractional exponent. Also, changed all \frac to \dfrac. Breathing room.

 

[lookagain]

\(\displaystyle \frac{x^2(x^2+9)^{-1/2}-\sqrt{x^2+9}}{x^2}\)

From the problem before I recognized that I should be focused on making radical expressions into fractional exponents.

\(\displaystyle > > > \frac{x^2(x^2+9)^{-1/2}-(x^2+9)^{-1/2}}{x^2} < < < \)

The above highlighted step is not correct. Note the sign on the second
fractional exponent.

If the top line (in this quote box) is correct, then the next step could be:


\(\displaystyle \dfrac{x^2(x^2 + 9)^{-1/2} - (x^2 + 9)^{1/2}}{x^2}\)

 

[BlBl]

The above highlighted step is not correct. Note the sign on the second
fractional exponent.

If the top line (in this quote box) is correct, then the next step could be:


\(\displaystyle \dfrac{x^2(x^2 + 9)^{-1/2} - (x^2 + 9)^{1/2}}{x^2}\)

Thanks for the catch. I've corrected the original post. It might've tripped me up if I were working solely from the display, but I was just trying to copy what I'd already written in the notebook. I c+p'd the first part of the expression and that's why that negative appeared. Still got the answer in the book though.

Alos, thank for showing me /dfrac. That makes things a lot easier to read.

 

[BlBl]

With "uglies(!)" like that, I find it's always better to simplify and re-arrange
as much as possible before "getting going" ; like this:

Let a = x^2 - 9 ; then:

1st term in numerator: 3 a^(1/3)

2nd term in numerator: (8x^2 / 3) / a^(2/3) : 2x * 4x * 1/3 = 8x^2 / 3 : OK?

denominator: a^(2/3)

So we have:
[3a^(1/3) - (8x^2 / 3) / a^(2/3)] / a^(2/3)

= [(3a - 8x^2 / 3) / a^(2/3)] / a^(2/3)

= (9a - 8x^2) / 3] / a^(4/3)

= (9a - 8x^2) / [3a^(4/3)]

Now subtitute a = x^2 - 9 back in...

Thanks for all that Dennis. It seems the habit of substituting and simplifying in that way simply hasn't become natural to me yet. I'll hit it again, and see what comes out.