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Math Help - Simplifying Expression 4

  1. #1
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    Simplifying Expression 4

    This is more difficult, I see familiar things (formulas basically behind numbers), but end with different result. How it is good to start simplifying it?

    Expression:
    \frac{4a^2-1}{(a^2-1)(a-1)}\div(\frac{a}{a^2-2a+1}-\frac{a}{(1-a)(a+1)}-\frac{2}{a+1})-\frac{1}{2}

    Answer:
    a

    Things what I see:
    Formula: a^2-b^2=(a+b)(a-b)

    4a^2-1=(2a+1)(2a-1)
    same thing with: (a^2-1)(a-1)

    (a^2-1)(a-1)=(a+1)(a-1)(a-1)

    I am not sure about this one:
    (a+1)(a-1)(a-1)=(a+1)(a-1)^2

    Going inside parenthesis:
    Formula: (a-b)^2

    a^2-2a+1=(a-1)^2

    Common denominator in my opinion is:
    \frac{a(a^2-1)-a(a-1)^2-2(a^2-1)(a-1)}{(a-1)^2(a^2-1)}

    Ready to divide (which is multiplication with opposite number):
    \frac{(2a+1)(2a-1)*(a-1)^2(a^2-1)}{(a+1)(a-1)^2*a(a^2-1)-a(a-1)^2-2(a^2-1)(a-1)}

    I know it looks crazy.
    \frac{(2a+1)(2a-1)*(a-1)}{a(a^2-1)-a(a-1)^2-2(a^2-1)(a-1)}

    I calculated it and got as final result:
    \frac{1}{2a}
    Interesting, but when I set a = 1 my simplification and textbook answer give different results. I went wrong, tips how I should approach it?
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  2. #2
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    Quote Originally Posted by Sooda View Post
    This is more difficult, I see familiar things (formulas basically behind numbers), but end with different result. How it is good to start simplifying it?

    Expression:
    \frac{4a^2-1}{(a^2-1)(a-1)}\div(\frac{a}{a^2-2a+1}-\frac{a}{(1-a)(a+1)}-\frac{2}{a+1})-\frac{1}{2}

    Answer:
    a

    ...
    \frac{4a^2-1}{(a^2-1)(a-1)}\div(\frac{a}{a^2-2a+1}-\frac{a}{(1-a)(a+1)}-\frac{2}{a+1})-\frac{1}{2}

    becomes

    \frac{4a^2-1}{(a^2-1)(a-1)}\div(\frac{a}{a^2-2a+1}+\frac{a}{(a-1)(a+1)}-\frac{2}{a+1})-\frac{1}{2}

    \frac{4a^2-1}{(a^2-1)(a-1)}\div \left(\frac{a^2+a+a^2-a-2(a^2-2a+1)}{(a^2-2a+1)(a+1)}\right)-\frac{1}{2}

    \frac{4a^2-1}{(a+1)(a-1)^2}\div \left(\frac{2(2a-1)}{(a-1)^2 (a+1)}\right)-\frac{1}{2}

    \frac{4a^2-1}{(a+1)(a-1)^2}\cdot \left(\frac{(a-1)^2 (a+1)}{2(2a-1)}\right)-\frac{1}{2}

    \dfrac{2a+1}2-\dfrac12=a+\dfrac12-\dfrac12=a
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Sooda View Post
    This is more difficult, I see familiar things (formulas basically behind numbers), but end with different result. How it is good to start simplifying it?

    Expression:
    \frac{4a^2-1}{(a^2-1)(a-1)}\div(\frac{a}{a^2-2a+1}-\frac{a}{(1-a)(a+1)}-\frac{2}{a+1})-\frac{1}{2}

    Answer:
    a

    Things what I see:
    Formula: a^2-b^2=(a+b)(a-b)

    4a^2-1=(2a+1)(2a-1)
    same thing with: (a^2-1)(a-1)

    (a^2-1)(a-1)=(a+1)(a-1)(a-1)

    I am not sure about this one:
    (a+1)(a-1)(a-1)=(a+1)(a-1)^2

    Going inside parenthesis:
    Formula: (a-b)^2

    a^2-2a+1=(a-1)^2

    Common denominator in my opinion is:
    \frac{a(a^2-1)-a(a-1)^2-2(a^2-1)(a-1)}{(a-1)^2(a^2-1)}

    Ready to divide (which is multiplication with opposite number):
    \frac{(2a+1)(2a-1)*(a-1)^2(a^2-1)}{(a+1)(a-1)^2*a(a^2-1)-a(a-1)^2-2(a^2-1)(a-1)}

    I know it looks crazy.
    \frac{(2a+1)(2a-1)*(a-1)}{a(a^2-1)-a(a-1)^2-2(a^2-1)(a-1)}

    I calculated it and got as final result:
    \frac{1}{2a}
    Interesting, but when I set a = 1 my simplification and textbook answer give different results. I went wrong, tips how I should approach it?
    ok
    the denominator

    (\frac{a}{a^2-2a+1}-\frac{a}{(1-a)(a+1)}-\frac{2}{a+1})

    \frac{a}{(1-a)^2}-\frac{a}{(1-a)(a+1)}-\frac{2}{(a+1)}

    the common denominator is (1-a)^2(a+1) note : (a-1)^2=(1-a)^2

    \frac{a(1+a)-a(1-a)-2(1-a)^2}{(1-a)^2(a+1)}

    \frac{a+a^2-(a-a^2)-2(a^2-2a+1)}{(1-a)^2(a+1)} simplify it

    \frac{4a-2}{(1-a)^2(a+1)} ok now we have

    \frac{4a^2-1}{(a^2-1)(a-1)}\div \frac{4a-2}{(1-a)^2(a+1)}-\frac{1}{2}

    \frac{4a^2-1}{(a^2-1)(a-1)} \times \frac{(1-a)^2(a+1)}{4a-2}-\frac{1}{2}

    \frac{(2a-1)(2a+1)}{2(2a-1)}-\frac{1}{2}

    \frac{2a+1}{2}-\frac{1}{2}=a
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  4. #4
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    How you know a^2-2a+1 has (a-1) already in it?

    EDIT:
    Where I can learn about this operation:
    -\frac{a}{(1-a)(1+a)}=\frac{a}{(a-1)(a+1)}
    ?

    I am confused how arguments move in denominator, by what rule?

    EDIT2:
    I have never seen something like that:
    <br />
(a-1)^2=(1-a)^2<br />

    Is it only true with <br />
(a-1)^2=(1-a)^2<br />
or would (4-3)^2=(3-4)^2 ?
    Last edited by Sooda; July 10th 2009 at 10:17 AM.
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  5. #5
    MHF Contributor Amer's Avatar
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    How you know a^2-2a+1 has (a-1) already in it?

    EDIT:
    Where I can learn about this operation:
    -\frac{a}{(1-a)(1+a)}=\frac{a}{(a-1)(a+1)}
    ?
    do you know how to find roots of quartic polynomials ??


    for the second question

    -\frac{a}{(1-a)(1+a)}=\frac{-a}{(1-a)(1+a)}=\frac{a}{-(1-a)(1+a)} since \frac{-a}{b} = \frac{a}{-b}

    \frac{a}{(a-1)(a+1)}


    EDIT2:
    I have never seen something like that:
    <br />
(a-1)^2=(1-a)^2<br />

    Is it only true with <br />
(a-1)^2=(1-a)^2<br />
or would (4-3)^2=(3-4)^2 ?
    try to expand it you will see that it is the same if you do not want to expand it

    you know that any negative with square power will be positive

    (-a)^2 = (-a)(-a)=a^2

    suppose that (1-a) positive number then (a-1) will be negative but if you take the square power the two number will be the same

    -3^2=9 , 3^2=9

    is there anything not clear ??
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  6. #6
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    Tallinn, Estonia
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    I overlooked something obvious here, afterwards I ask: "How you know
    a^2-2a+1<br />
has (a-1)<br />
in it? and I wrote in my first post:
    Going inside parenthesis:

    Formula: (a-b)^2<br />

    a^2-2a+1=(a-1)^2<br />


    > do you know how to find roots of quartic polynomials ??

    I am going to google about it. I have forgotten it.

    Everything makes sense when you typed it out. I keep low profile until I get smarter.
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