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    LynnM's Avatar
    LynnM Posts: 34, Reputation: 2
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    #1

    Aug 25, 2010, 03:32 PM
    Simplify the given expressions and express all answers with positive exponents
    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}
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Aug 26, 2010, 07:24 AM

    Do you mean this?



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



    Then, you can put it this way:



    1.



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



    Now, use the product rule, that is



    So, this becomes:



    Simplify the fractions;





    Which becomes:



    Now simplify further by factorising the numerator by -y:







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

    2.



    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.



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



    Simplify the terms in brackets and put the whole thing into a more 'aesthetic' way.
    LynnM's Avatar
    LynnM Posts: 34, Reputation: 2
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    #3

    Aug 26, 2010, 10:08 AM

    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.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Aug 26, 2010, 10:12 AM

    You can simplify to:

    LynnM's Avatar
    LynnM Posts: 34, Reputation: 2
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    #5

    Aug 26, 2010, 10:28 AM

    Hmmm, okay, can you explain how to factor in question #3. Math is not my forte, especially factoring. Sorry:)
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #6

    Aug 26, 2010, 10:39 AM


    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:



    It's like:



    Since this is division, you can subtract the powers:



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



    Which becomes:



    Is it okay? :)

    Now, you can further simplify :)
    LynnM's Avatar
    LynnM Posts: 34, Reputation: 2
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    #7

    Aug 26, 2010, 11:15 AM

    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:(
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #8

    Aug 27, 2010, 08:21 AM

    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:



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



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



    or expand to give:

    LynnM's Avatar
    LynnM Posts: 34, Reputation: 2
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    #9

    Aug 30, 2010, 10:05 AM

    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!

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