Results 1 to 8 of 8

Math Help - Solve the equation

  1. #1
    Senior Member
    Joined
    Nov 2007
    Posts
    329

    Solve the equation

    Solve the equation: a^2b^2+b^2c^2+c^2a^2+a^2+b^2+c^2+4(a+b+c)+12=6abc+  4(ab+bc+ca)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,698
    Thanks
    454
    Quote Originally Posted by james_bond View Post
    Solve the equation: a^2b^2+b^2c^2+c^2a^2+a^2+b^2+c^2+4(a+b+c)+12=6abc+  4(ab+bc+ca)
    solve for what?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Nov 2007
    Posts
    329
    Quote Originally Posted by skeeter View Post
    solve for what?
    For a, b and c. I've solved for the case a=b=c and got two solutions. I'm pretty sure that there are no other solutions but I can't prove it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2007
    Posts
    329
    So if a=b=c the equation becomes:
    3a^4+3a^2+12a+12=6a^3+12a^2\Leftrightarrow a^4-2a^3-3a^2+4a+4=0\Leftrightarrow  (-2 + a)^2 (1 + a)^2=0
    So a=b=c=2 and a=b=c=-1 are solutions.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Nov 2007
    Posts
    329
    No ideas how to prove that there are no other solutions? Maybe this equation is an inequality and equality holds iff a=b=c. If so what's the inequality and how to prove it?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    The equation can be written as

    (ab-c-2)^2+(ac-b-2)^2+(bc-a-2)^2=0

    Then

    \left\{\begin{array}{lll}ab-c=2\\ac-b=2\\bc-a=2\end{array}\right.

    Can you solve the system?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Nov 2007
    Posts
    329
    Thanks for your help but unfortunately I can't solve the equation system. I got ab+b=2 and similarly the other two but I can't get further. Could you help me out, please? Thank you!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    Quote Originally Posted by red_dog View Post

    \left\{\begin{array}{lll}ab-c=2\\ac-b=2\\bc-a=2\end{array}\right.
    Substract the second equation from the first:

    (b-c)(a+1)=0

    1. If b=c, replace c with b in the first and the third equation:

    \left\{\begin{array}{ll}ab-b=2\\b^2-a=2\end{array}\right.

    From the second equation a=b^2-2

    Replace a in the first equation: b^3-3b-2=0\Leftrightarrow (b+1)^2(b-2)=0

    If b=-1\Rightarrow a=-1, \ c=-1

    If b=2\Rightarrow a=2, \ c=2

    2. If a=-1, replace a in the first and the third equation:

    \left\{\begin{array}{ll}b+c=-2\\bc=1\end{array}\right.\Rightarrow b=c=-1
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Solve Differential equation for the original equation
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: February 21st 2011, 01:24 PM
  2. Solve the equation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: August 9th 2010, 11:48 AM
  3. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: July 16th 2010, 10:29 PM
  4. solve equation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 29th 2008, 08:12 AM
  5. Replies: 13
    Last Post: May 19th 2008, 08:56 AM

Search Tags


/mathhelpforum @mathhelpforum