Results 1 to 6 of 6

Math Help - proof for an equation

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    4

    proof for an equation

    I have this equation here:

    x^2 + y^2 + z^2 = xy + xz + yz

    I need to somehow prove that the equation is true only when all three variables are equal, i.e., x = y = z.

    I would appreciate any advice or suggestions on how to accomplish this.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by gfreeman75 View Post
    I have this equation here:

    x^2 + y^2 + z^2 = xy + xz + yz

    I need to somehow prove that the equation is true only when all three variables are equal, i.e., x = y = z.

    I would appreciate any advice or suggestions on how to accomplish this.
    Well, I'll probably get jumped on here but here goes .......

    The equation can be re-arranged in two ways:

    x(x - y) + y(y - z) + z(z - x) = 0 .... (1)

    x(x - z) + y(y - x) + z(z - y) = 0 ....(2)

    Compare and equate coefficients of x, y and z:

    x - y = x - z => y = z.

    y - z = y - x => x = z.

    Therefore x = y = z is the only solution.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by gfreeman75 View Post
    I have this equation here:

    x^2 + y^2 + z^2 = xy + xz + yz

    I need to somehow prove that the equation is true only when all three variables are equal, i.e., x = y = z.

    I would appreciate any advice or suggestions on how to accomplish this.
    x^2 + y^2 + z^2 = xy + xz + yz

    \Rightarrow 2(x^2 + y^2 + z^2) = 2(xy + xz + yz)

    \Rightarrow 2(x^2 + y^2 + z^2) - 2(xy + xz + yz) = 0

    \Rightarrow (x^2 + y^2 - 2xy)+(x^2 + z^2 - 2xz) + (z^2 + y^2 - 2zy) = 0

    \Rightarrow (x-y)^2+(x-z)^2 + (z-y)^2 = 0

    \text{This forces }x=y=z.

    Quote Originally Posted by Mr.F
    Well, I'll probably get jumped on here but here goes .......
    Mr F, can you please explain the validity of your method?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by Isomorphism View Post
    x^2 + y^2 + z^2 = xy + xz + yz

    \Rightarrow 2(x^2 + y^2 + z^2) = 2(xy + xz + yz)

    \Rightarrow 2(x^2 + y^2 + z^2) - 2(xy + xz + yz) = 0

    \Rightarrow (x^2 + y^2 - 2xy)+(x^2 + z^2 - 2xz) + (z^2 + y^2 - 2zy) = 0

    \Rightarrow (x-y)^2+(x-z)^2 + (z-y)^2 = 0

    \text{This forces }x=y=z.


    Mr F, can you please explain the validity of your method?
    Very succinct. It took me at least twice as many lines when I tried working it out.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2008
    Posts
    4
    That's a pretty elegant solution to my problem, Isomorphism. Thank you.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Isomorphism View Post
    x^2 + y^2 + z^2 = xy + xz + yz

    \Rightarrow 2(x^2 + y^2 + z^2) = 2(xy + xz + yz)

    \Rightarrow 2(x^2 + y^2 + z^2) - 2(xy + xz + yz) = 0

    \Rightarrow (x^2 + y^2 - 2xy)+(x^2 + z^2 - 2xz) + (z^2 + y^2 - 2zy) = 0

    \Rightarrow (x-y)^2+(x-z)^2 + (z-y)^2 = 0

    \text{This forces }x=y=z.


    Mr F, can you please explain the validity of your method?
    Well, I guess I'm saying that if you've got

    ax + by + cz = 0 .... (1)

    cx + ay + bz = 0 ....(2)

    then a = b = c if x, y and z are not equal to zero.

    A bit lame, I guess.
    Last edited by mr fantastic; April 28th 2008 at 03:25 PM. Reason: Removed "= 0", which was really really lame
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Proof hyperbolic equation
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 11th 2011, 07:42 PM
  2. Proof an equation
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: December 21st 2010, 11:24 AM
  3. Characteristic equation proof
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 6th 2009, 08:37 PM
  4. Please Help me proof the equation!
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: January 11th 2009, 01:40 PM
  5. Quadratic equation proof
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 4th 2008, 01:32 PM

Search Tags


/mathhelpforum @mathhelpforum