Results 1 to 4 of 4

Thread: How to divide GCF through...

  1. April 1st 2011, 03:20 PM #1
    kungfumonkey
    kungfumonkey is offline
    Newbie
    Joined
    Apr 2011
    Posts
    6

    Red face How to divide GCF through...

    Hi and thanks in advance...

    If I have this:
    % MathType!MTEF!2!1!+-<br /> % faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj<br /> % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x<br /> % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq<br /> % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa<br /> % aiaabeqaamaabaabaaGcbaGaaGOmaiaadIhacaGGOaGaamiEam aaCa<br /> % aaleqabaGaaGOmaaaakiabgUcaRiaaiMdacaGGPaWaaWbaaSqa beaa<br /> % caaIXaGaai4laiaaikdaaaGccqGHRaWkcaWG4bWaaWbaaSqabe aaca<br /> % aIZaaaaOGaaiikaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGH RaWk<br /> % caaI5aGaaiykamaaCaaaleqabaGaaiylaiaaigdacaGGVaGaaG Omaa<br /> % aaaaa!41E3!<br /> \[<br /> 2x(x^2 + 9)^{1/2} + x^3 (x^2 + 9)^{ - 1/2} <br /> \]<br />

    I understand that my GCF is :
    % MathType!MTEF!2!1!+-<br /> % faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj<br /> % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x<br /> % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq<br /> % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa<br /> % aiaabeqaamaabaabaaGcbaGaamiEaiaacIcacaWG4bWaaWbaaS qabe<br /> % aacaaIYaaaaOGaey4kaSIaaGyoaiaacMcadaahaaWcbeqaaiab gkHi<br /> % TiaaigdacaGGVaGaaGOmaaaaaaa!3741!<br /> \[<br /> x(x^2 + 9)^{ - 1/2} <br /> \]<br />

    What I don't understand is how to divide through by my GCF. What is throwing me off is that I don't know what to do with the -1/2 and 1/2 exponents... Can anybody please help me understand this? My professor is too fast and I'm the only idiot in the class! No other help is available to me until Monday and this is killing me! Thanks again, in advance for your time and help!
    Follow Math Help Forum on Facebook and Google+
  2. April 1st 2011, 04:02 PM #2
    pickslides
    pickslides is offline
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    26
    \displaystyle 2x(x^2 + 9)^{1/2} + x^3 (x^2 + 9)^{ - 1/2}

    \displaystyle \frac{2x(x^2 + 9)^{1/2}}{x(x^2 + 9)^{ - 1/2}} + \frac{x^3 (x^2 + 9)^{ - 1/2}}{x(x^2 + 9)^{ - 1/2}}

    Using index law \displaystyle a^m \div a^n = a^{m-n}

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

    Finish him from here?
    Follow Math Help Forum on Facebook and Google+
  3. April 1st 2011, 04:02 PM #3
    topsquark
    topsquark is offline
    Moderator
    topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,080
    Thanks
    153
    Awards
    1
    MHF Donor
    Quote Originally Posted by kungfumonkey View Post
    Hi and thanks in advance...

    If I have this:
    % MathType!MTEF!2!1!+-<br /> % faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj<br /> % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x<br /> % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq<br /> % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa<br /> % aiaabeqaamaabaabaaGcbaGaaGOmaiaadIhacaGGOaGaamiEam aaCa<br /> % aaleqabaGaaGOmaaaakiabgUcaRiaaiMdacaGGPaWaaWbaaSqa beaa<br /> % caaIXaGaai4laiaaikdaaaGccqGHRaWkcaWG4bWaaWbaaSqabe aaca<br /> % aIZaaaaOGaaiikaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGH RaWk<br /> % caaI5aGaaiykamaaCaaaleqabaGaaiylaiaaigdacaGGVaGaaG Omaa<br /> % aaaaa!41E3!<br /> \[<br /> 2x(x^2 + 9)^{1/2} + x^3 (x^2 + 9)^{ - 1/2} <br /> \]<br />

    I understand that my GCF is :
    % MathType!MTEF!2!1!+-<br /> % faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj<br /> % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x<br /> % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq<br /> % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa<br /> % aiaabeqaamaabaabaaGcbaGaamiEaiaacIcacaWG4bWaaWbaaS qabe<br /> % aacaaIYaaaaOGaey4kaSIaaGyoaiaacMcadaahaaWcbeqaaiab gkHi<br /> % TiaaigdacaGGVaGaaGOmaaaaaaa!3741!<br /> \[<br /> x(x^2 + 9)^{ - 1/2} <br /> \]<br />

    What I don't understand is how to divide through by my GCF. What is throwing me off is that I don't know what to do with the -1/2 and 1/2 exponents... Can anybody please help me understand this? My professor is too fast and I'm the only idiot in the class! No other help is available to me until Monday and this is killing me! Thanks again, in advance for your time and help!
    The 1/2 exponents are merely square roots. For example:
    \displaystyle 2x \sqrt{x^2 + 9} + \frac{x^3}{\sqrt{x^2 + 9}}

    Does this help?

    -Dan
    Follow Math Help Forum on Facebook and Google+
  4. April 1st 2011, 04:15 PM #4
    kungfumonkey
    kungfumonkey is offline
    Newbie
    Joined
    Apr 2011
    Posts
    6

    Thumbs up

    Thanks, you folks rule!

    I ended up with this thanks to your advice:
    % MathType!MTEF!2!1!+-<br /> % faaagaart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedm vETj<br /> % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpe pC0x<br /> % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq<br /> % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa<br /> % aiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIZaGaamiEaiaacI caca<br /> % WG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOnaiaacMca aeaa<br /> % daGcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkca aI5a<br /> % aaleqaaaaaaaa!3877!<br /> \[<br /> \frac{{3x(x^2 + 6)}}{{\sqrt {x^2 + 9} }}<br /> \]<br />

    And it seems that's the correct solution! THANK YOU SO MUCH!!!

    Follow Math Help Forum on Facebook and Google+
« how to determin number of combinations | solution set for y = 2x - 5 »

Similar Math Help Forum Discussions

  1. Divide
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 2nd 2010, 02:52 PM
  2. how to divide an arc by...
    Posted in the Geometry Forum
    Replies: 3
    Last Post: February 1st 2009, 07:10 AM
  3. divide
    Posted in the Algebra Forum
    Replies: 4
    Last Post: January 16th 2009, 05:51 AM
  4. help divide
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 6th 2007, 11:37 AM
  5. Divide 2Pi into 4
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: June 27th 2005, 08:13 PM

Search Tags


/mathhelpforum @mathhelpforum