Hi. I've almost finished my assignment but there are three questions that I just can't figure out. If someone could provide the steps to finding the answer, I would really appreciate it. (One thing that really confuses me is when a part of the equation looks like this: x^2(2x-1)^{1/2}. Are we allowed to multiply the first part without doing anything with the exponent?) Anyway, here are the questions:

#1: [y^{3/8} (y^{5/8} - y^{13/8})] / [y^{1/2} (y^{1/2} - y^{-1/2})]

#2: [(a^{1/2} - a^{-1/2})^{2} +4]^{1/2}

#3: x^{2} (2x - 1)^{-1/2} + 2x (2x - 1)^{1/2}

Do you mean this?

x^2(2x-1)^{\frac12}

and using the distributive law?
This you cannot because the terms should be of the same power. For example, if it were:

x^2(2x-1)^2

Then, you can put it this way:

x^2(2x-1)^2 = [x(2x-1)]^2 = (2x^2 - x)^2

1.

\Large{\frac{y^{\frac38} \(y^{\frac58} - y^{\frac{13}{8}}\)}{y^{\frac12} \(y^{\frac12} - y^{-\frac12}\)}}

Expand the numerator and the denominator, leave the powers as they are, don't simplify now.

=\Large{\frac{y^{\frac38}.y^{\frac58} - y^{\frac38}.y^{\frac{13}{8}}}{y^{\frac12}.y^{\frac 12} - y^{\frac12}.y^{-\frac12}}}

Now, use the product rule, that is

a^xa^y = a^{(x+y)}

So, this becomes:

=\Large{\frac{y^{\frac38 + \frac58} - y^{\frac38 + \frac{13}{8}}}{y^{\frac12 + \frac12} - y^{\frac12 - \frac12}}

Simplify the fractions;

=\Large{\frac{y^{\frac88} - y^{\frac{16}{8}}}{y^{\frac22} - y^{\frac02}}

=\Large{\frac{y^1 - y^2}{y^1 - y^0}

Which becomes:

=\Large{\frac{y - y^2}{y - 1}

Now simplify further by factorising the numerator by -y:

=\Large{\frac{-y(-1 +y)}{y - 1}

=\Large{\frac{-y(y-1)}{y - 1}

=-y

Can you try the others now and post what you get? :)

2.

((a^{\frac12} - a^{-\frac12})^2 +4)^{\frac12}

Expand the quadratic first.
Convert everything into a single fraction.
Factorise the numerator.
Then, insert the power (1/2) into the brackets.
Now, you can separate the improper fraction into two terms.

3.

x^2(2x - 1)^{-\frac12} + 2x (2x - 1)^{\frac12}

Factorise first. You can factorise x and (2x-1)^(-1/2).
Remember that

\frac{(2x - 1)^{\frac12}}{(2x - 1)^{-\frac12}} = (2x-1)^{\frac12--\frac12} = (2x-1)^{1} = (2x-1)

Simplify the terms in brackets and put the whole thing into a more 'aesthetic' way.

Oh my goodness! I have a big 'note-to-self' saying 'DON'T forget about factoring' and that's just what I didn't do. I forgot all about it. Anyway, here's what I've done for #2:

#2: [(a^{1/2} - a^{-1/2})^{2} +4]^{1/2}

[((a^{1/2}-a^{-1/2})(a^{1/2}-a^{-1/2}))+4]^{1/2}

[(a-1-1+a^{-1})+4]^{1/2}

[(a-2+(1/a))+4]^{1/2}

[(a^{2} + 2a + 1) / a]^{1/2}

(a+1^{1/2 + 1/2}) / a^{1/2}

(a+1) / a^{1/2}

I will post #3 right away.

You can simplify to:

\frac{a+1}{a^{\frac12}} = \frac{a}{\sqrt{a}} + \frac{1}{\sqrt{a}} = \sqrt{a} + \frac{1}{\sqrt{a}}

Hmmm, okay, can you explain how to factor in question #3. Math is not my forte, especially factoring. Sorry:)

x^2(2x - 1)^{-\frac12} + 2x (2x - 1)^{\frac12}

You know that you have x as common factor.

And you have (2x-1) which are in both terms, but not in the same powers... so, take the lowest power. That's (2x-1)^-1/2

Now, this thing, I told you earlier:

\frac{(2x - 1)^{\frac12}}{(2x - 1)^{-\frac12}} = (2x-1)

It's like:

\frac{a^2}{a^3}

Since this is division, you can subtract the powers:

\frac{a^2}{a^3} = a^{2-3} = a^{-1} = \frac{1}{a}

The same thing applies. Since you are factoring (2x-1)^-1/2, you are dividing by this term. Or:

x^2(2x - 1)^{-\frac12} + 2x (2x - 1)^{\frac12} = x(2x-1)^{-\frac12}\[\frac{x^2(2x - 1)^{-\frac12}}{x(2x-1)^{-\frac12}} + \frac{2x (2x - 1)^{\frac12}}{x(2x-1)^{-\frac12}}\]

Which becomes:

x(2x-1)^{-\frac12}[x + 2 (2x - 1)]

Is it okay? :)

Now, you can further simplify :)

Sorry, this may take a while:( Can you tell me if I am at least on the right track.

[x / (2x-1)^{-1/2}] * [x + (4x-2)]

[x^{2} + (4x^{2} - 2x)] / (2x-1)^{1/2}

This question has got me stumped:(

Um... I guess your first line was a typo from your part, which read in fact:

[x / (2x-1)^{1/2}] * [x + (4x-2)]

instead of

[x / (2x-1)^{-1/2}] * [x + (4x-2)]

This done, here is what it is:

\frac{x}{\sqrt{2x-1}}\( x + 4x - 2\)

You don't need to multiply by x now. Simplify that in the brackets:

\frac{x}{\sqrt{2x-1}}\(5x - 2\)

Now, if you want, you can put it like:

\frac{x(5x - 2)}{\sqrt{2x-1}}

or expand to give:

\frac{5x^2 - 2x}{\sqrt{2x-1}}

I have been away from my math for a few days so everything is a little fuzzy right now. I think I have a pretty good understanding of this question now. I will try a few more like it to make sure. Thank you so much for you help... and your patience!