Results 1 to 4 of 4

Thread: Solving Equations in Quadratic Form

  1. #1
    Newbie
    Joined
    Sep 2006
    Posts
    13

    Solving Equations in Quadratic Form

    Hi I'm kinda confused by these equations.

    1.) 3[ (x+3) / (2x-1) ]^2 - 4(x+3) / (2x-1) + 1 = 0

    2.) 36x^4/3 - 25x^2/3 + 4 = 0

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Sep 2008
    Posts
    27
    3$\displaystyle (x+3 / 2x-1)^2 - 4(x+3 / 2x-1) + 1$
    = $\displaystyle 3a^2 - 4a + 1 $ { replacing x$\displaystyle +3/2x-1) by a $}
    = $\displaystyle 3a^2 - 3a - a + 1$
    =
    $\displaystyle 3a(a - 1) - 1(a-1)$= $\displaystyle (3a-1)(a-1)$
    = $\displaystyle {3(x+3 / 2x-1) - 1}{3(x+3 / 2x-1) - 1} $
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    Posts
    27
    $\displaystyle
    36x^4/3 - 25x^2/3 + 4 = 0
    $
    Or $\displaystyle 36a^2 - 25a + 4 = 0 $ {subtituting $\displaystyle x^2/3$ by a}
    Or $\displaystyle 36a^2 - 16a -9a + 4=0$
    Or $\displaystyle 4a(9a-4) - 1(9a-4)=0$
    Or (4a-1)(9a-4)=0
    Or $\displaystyle a=1/4, 4/9$
    Or $\displaystyle x^2/3 = 1/4 , 4/9$
    Or $\displaystyle x^1/3=1/2 , 2/3$
    Or $\displaystyle x = 1/8 ,8/27. $
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2008
    Posts
    27
    You're welcome!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Sep 5th 2011, 07:13 AM
  2. Solving Quadratic Equations in Factored Form
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Aug 10th 2010, 04:19 PM
  3. Solving Equations in Quadratic Form
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Oct 8th 2008, 08:54 AM
  4. Replies: 1
    Last Post: Jun 12th 2008, 09:30 PM
  5. Replies: 14
    Last Post: May 30th 2008, 06:10 AM

Search Tags


/mathhelpforum @mathhelpforum