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Thread: How do I simplify this?

  1. #1
    Newbie
    Joined
    Apr 2011
    Posts
    6

    How do I simplify this?

    Hi folks!

    I am not sure of what to do next here:

    Numerator: $\displaystyle % MathType!MTEF!2!1!+-
    % faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj
    % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x
    % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq
    % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa
    % aiaabeqaamaabaabaaGcbaGaeyOeI0YaaSaaaeaacaaIXaaaba GaaG
    % OmaaaacaWG4bWaaWbaaSqabeaacqGHsislcaaIXaGaai4laiaa ikda
    % aaGccaWG5bWaaWbaaSqabeaacaaIXaGaai4laiaaikdaaaGccq GHsi
    % slcaGGBbGaeyOeI0YaaSaaaeaacaaIXaaabaGaaGOmaaaacaWG 4bWa
    % aWbaaSqabeaacaaIXaGaai4laiaaikdaaaGccaWG5bWaaWbaaS qabe
    % aacqGHsislcaaIXaGaai4laiaaikdaaaGcdaqadaqaamaalaaa baGa
    % eyOeI0IaamiEamaaCaaaleqabaGaaGymaiaac+cacaaIYaaaaa Gcba
    % GaamyEamaaCaaaleqabaGaaGymaiaac+cacaaIYaaaaaaaaOGa ayjk
    % aiaawMcaaiaac2faaaa!4E3E!
    \[
    { - \frac{1}
    {2}x^{ - 1/2} y^{1/2} - [ - \frac{1}
    {2}x^{1/2} y^{ - 1/2} \left( {\frac{{ - x^{1/2} }}
    {{y^{1/2} }}} \right)]}
    \]
    $
    All over this denominator: $\displaystyle % MathType!MTEF!2!1!+-
    % faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj
    % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x
    % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq
    % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa
    % aiaabeqaamaabaabaaGcbaWaaeWaaeaacaWG5bWaaWbaaSqabe aaca
    % aIXaGaai4laiaaikdaaaaakiaawIcacaGLPaaadaahaaWcbeqa aiaa
    % ikdaaaaaaa!33E3!
    \[
    {\left( {y^{1/2} } \right)^2 }
    \]
    $
    (too much latex for the post) How should I approach this? Is there something I should be checking for first? Here's the answer, but I can't figure out how to get from the above to the answer:

    $\displaystyle % MathType!MTEF!2!1!+-
    % faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj
    % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x
    % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq
    % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa
    % aiaabeqaamaabaabaaGcbaWaaSaaaeaacqGHsislcaaIXaaaba GaaG
    % OmaiaadMhadaahaaWcbeqaaiaaikdaaaGcdaGcaaqaaiaadIha aSqa
    % baaaaaaa!338F!
    \[
    \frac{{ - 1}}
    {{2y^2 \sqrt x }}
    \]
    $
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  2. #2
    Member
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    Mar 2011
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    Awetuouncsygg
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    Well, for simplicity note a=$\displaystyle x^\frac{1}{2}$ and b=$\displaystyle y^\frac{1}{2}$. So you have: $\displaystyle \frac{-\frac{1}{2}a^{-1}b-\frac{1}{2}ab^{-1}\cdot \frac{a}{b}}{b^2}$

    Use that $\displaystyle a^{-1}=\frac{1}{a}$. Hope you can solve it now.

    Edit: You sure wrote that correctly/ completely? Because $\displaystyle a^3+b^3$ should be equal with 1 which is not necessary true.
    Last edited by veileen; Apr 3rd 2011 at 12:50 AM.
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  3. #3
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    456
    Quote Originally Posted by kungfumonkey View Post
    Hi folks!

    I am not sure of what to do next here:

    Numerator: $\displaystyle % MathType!MTEF!2!1!+-
    % faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj
    % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x
    % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq
    % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa
    % aiaabeqaamaabaabaaGcbaGaeyOeI0YaaSaaaeaacaaIXaaaba GaaG
    % OmaaaacaWG4bWaaWbaaSqabeaacqGHsislcaaIXaGaai4laiaa ikda
    % aaGccaWG5bWaaWbaaSqabeaacaaIXaGaai4laiaaikdaaaGccq GHsi
    % slcaGGBbGaeyOeI0YaaSaaaeaacaaIXaaabaGaaGOmaaaacaWG 4bWa
    % aWbaaSqabeaacaaIXaGaai4laiaaikdaaaGccaWG5bWaaWbaaS qabe
    % aacqGHsislcaaIXaGaai4laiaaikdaaaGcdaqadaqaamaalaaa baGa
    % eyOeI0IaamiEamaaCaaaleqabaGaaGymaiaac+cacaaIYaaaaa Gcba
    % GaamyEamaaCaaaleqabaGaaGymaiaac+cacaaIYaaaaaaaaOGa ayjk
    % aiaawMcaaiaac2faaaa!4E3E!
    \[
    { - \frac{1}
    {2}x^{ - 1/2} y^{1/2} - [ - \frac{1}
    {2}x^{1/2} y^{ - 1/2} \left( {\frac{{ - x^{1/2} }}
    {{y^{1/2} }}} \right)]}
    \]
    $
    All over this denominator: $\displaystyle % MathType!MTEF!2!1!+-
    % faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj
    % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x
    % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq
    % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa
    % aiaabeqaamaabaabaaGcbaWaaeWaaeaacaWG5bWaaWbaaSqabe aaca
    % aIXaGaai4laiaaikdaaaaakiaawIcacaGLPaaadaahaaWcbeqa aiaa
    % ikdaaaaaaa!33E3!
    \[
    {\left( {y^{1/2} } \right)^2 }
    \]
    $
    (too much latex for the post) How should I approach this? Is there something I should be checking for first? Here's the answer, but I can't figure out how to get from the above to the answer:

    $\displaystyle % MathType!MTEF!2!1!+-
    % faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj
    % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x
    % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq
    % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa
    % aiaabeqaamaabaabaaGcbaWaaSaaaeaacqGHsislcaaIXaaaba GaaG
    % OmaiaadMhadaahaaWcbeqaaiaaikdaaaGcdaGcaaqaaiaadIha aSqa
    % baaaaaaa!338F!
    \[
    \frac{{ - 1}}
    {{2y^2 \sqrt x }}
    \]
    $
    is this what you wrote there ?

    $\displaystyle \displaystyle \frac{-\frac {\sqrt{y}}{2\sqrt{x}} - \big{[} -\frac {\sqrt{x}}{2\sqrt{y}}\cdot (\frac{-\sqrt{x}}{\sqrt{y}})\big{]}}{y}$
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  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
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    Thanks
    848
    Hello, kungfumonkey!

    $\displaystyle \displaystyle \text{Simplify: }\;\frac{-\frac{1}{2}x^{-\frac{1}{2}}y^{\frac{1}{2}} - \bigg[-\frac{1}{2}x^{\frac{1}{2}}y^{-\frac{1}{2}}\left(-\frac{x^{\frac{1}{2}}}{y^{\frac{1}{2}}}\right)\big g]}{\left(y^{\frac{1}{2}}\right)^2} $

    $\displaystyle \text{Answer: }\:\dfrac{ - 1}{2y^2 \sqrt x }$ . How?

    $\displaystyle \displaystyle \text{We have: }\;\frac{-\frac{1}{2}x^{-\frac{1}{2}}y^{\frac{1}{2}}- \frac{1}{2}xy^{-1}}{y} \;\;=\;\;-\tfrac{1}{2}x^{-\frac{1}{2}}y^{-\frac{1}{2}} - \tfrac{1}{2}xy^{-2}$

    . . . . $\displaystyle \displaystyle =\;-\tfrac{1}{2}\left(\frac{1}{x^{\frac{1}{2}}y^{\frac {1}{2}}} + \frac{x}{y^2}\right) \;\;=\;\;-\tfrac{1}{2}\left(\frac{1}{x^{\frac{1}{2}}y^{\frac {1}{2}}}\cdot\frac{y^{\frac{3}{2}}}{y^{\frac{3}{2} }} + \frac{x}{y^2}\cdot\frac{x^{\frac{1}{2}}}{x^{\frac{ 1}{2}}}\right)$

    . . . . $\displaystyle \displaystyle =\;\;-\tfrac{1}{2}\left(\frac{y^{\frac{3}{2}}}{x^{\frac{ 1}{2}}y^2} + \frac{x^{\frac{3}{2}}}{x^{\frac{1}{2}}y^2}\right) \;\;=\;\;-\tfrac{1}{2}\left(\frac{x^{\frac{3}{2}} + y^{\frac{3}{2}}}{x^{\frac{1}{2}}y^2}\right) $

    . . . . $\displaystyle \displaystyle =\;\;-\frac{x\sqrt{x} + y\sqrt{y}}{2y^2\sqrt{x}} $


    As veileen pointsd out, their answer suggests that
    . . the numerator equals 1, which is not true.


    The numerator factors: .$\displaystyle x\sqrt{x} + y\sqrt{y} \;=\;(\sqrt{x} + \sqrt{y})(x - \sqrt{xy} + y)$
    . . which is of absolutely no help.

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  5. #5
    Newbie
    Joined
    Apr 2011
    Posts
    6
    Thanks for the help all! The solution provided is what my solution manual gives but it has been incorrect before, even on testing. I don't have any examples that match this format so I can't compare my work to anything in my course book.

    I'll follow up Monday with what my professor says.
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  6. #6
    Newbie
    Joined
    Apr 2011
    Posts
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    My professor said there was a mistake in the book. The correct solution was what Soroban posted. Thanks again gurus!
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