Results 1 to 6 of 6
Like Tree2Thanks
  • 1 Post By ChessTal
  • 1 Post By ChessTal

Math Help - Little problem I created for myself

  1. #1
    Member
    Joined
    Oct 2011
    Posts
    170
    Thanks
    3

    Little problem I created for myself

    I want to have some algebra expression so that once it is solved the answer becomes E=mc^2. I want it to be a little complicated though where the answer isn't obvious and you do actually have to reduces, cancel ect. Something like this for example. I know it doesn't mean anything but to give you an idea, once solved and everything cancels ect the solution is E=mc^2

    ac^2+\frac{E^2}{6-m^2} = -6-a \frac{1}{E^2}

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,842
    Thanks
    320
    Awards
    1

    Re: Little problem I created for myself

    Quote Originally Posted by uperkurk View Post
    I want to have some algebra expression so that once it is solved the answer becomes E=mc^2. I want it to be a little complicated though where the answer isn't obvious and you do actually have to reduces, cancel ect. Something like this for example. I know it doesn't mean anything but to give you an idea, once solved and everything cancels ect the solution is E=mc^2

    ac^2+\frac{E^2}{6-m^2} = -6-a \frac{1}{E^2}

    Thanks!
    Well, you could start with E = mc^2 and add a term to both sides, divide both sides by something, square both sides, etc. If you want to have real fun you could do something like...
    E = mc^2

    E + 3ac = mc^2 + 10ac^3
    or something and make them solve for a such that E = mc^2.

    Of course you could have some real fun and use E^2 = p^2c^2 + (mc^2)^2, where p is the momentum. That's the general form for E = mc^2.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2011
    Posts
    170
    Thanks
    3

    Re: Little problem I created for myself

    Thanks topsquark, I had something like that in mind but I wanted to make it a little more complicated and not make it apparent that the answer is E=mc^2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2010
    Posts
    52
    Thanks
    17

    Re: Little problem I created for myself

    Quote Originally Posted by uperkurk View Post
    I want to have some algebra expression so that once it is solved the answer becomes E=mc^2. I want it to be a little complicated though where the answer isn't obvious and you do actually have to reduces, cancel ect. Something like this for example. I know it doesn't mean anything but to give you an idea, once solved and everything cancels ect the solution is E=mc^2

    ac^2+\frac{E^2}{6-m^2} = -6-a \frac{1}{E^2}

    Thanks!
    As an exercise for someone you can ask him to solve for E and x(both real number) and for m \ne 0 and c \ne 0 the following:

    2{E^2} + {(mxc)^2} + {m^2}{c^4} - 2mxcE - 2{c^2}Em \le 0

    or even better the equivalent inequality:

    2{\left( {\frac{E}{c}} \right)^2} + {(mc)^2} + {(mx)^2} - 2\frac{{mxE}}{c} - 2mE \le 0
    Last edited by ChessTal; July 23rd 2013 at 03:45 AM.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2011
    Posts
    170
    Thanks
    3

    Re: Little problem I created for myself

    Quote Originally Posted by ChessTal View Post
    As an exercise for someone you can ask him to solve for E and x(both real number) and for m \ne 0 and c \ne 0 the following:

    2{E^2} + {(mxc)^2} + {m^2}{c^4} - 2mxcE - 2{c^2}Em \le 0

    or even better the equivalent inequality:

    2{\left( {\frac{E}{c}} \right)^2} + {(mc)^2} + {(mx)^2} - 2\frac{{mxE}}{c} - 2mE \le 0
    When I paste these into Wolfram it tells me it doesn't understand the equation and the closest answer it gives is c^4m^2
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jan 2010
    Posts
    52
    Thanks
    17

    Re: Little problem I created for myself

    Quote Originally Posted by uperkurk View Post
    When I paste these into Wolfram it tells me it doesn't understand the equation and the closest answer it gives is c^4m^2
    Well Wolfram Alpha is wrong since this equation is solvable and the solution is what you ask.... Just solve it yourself and forget Wolfram A.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 22
    Last Post: December 28th 2013, 04:44 PM
  2. Matlab help: User created functions
    Posted in the Math Software Forum
    Replies: 7
    Last Post: July 7th 2010, 07:21 PM
  3. Scaling in already created eq's in mathtype
    Posted in the LaTeX Help Forum
    Replies: 1
    Last Post: February 8th 2010, 07:45 AM
  4. what's wrong with my pdf created by texshop?
    Posted in the LaTeX Help Forum
    Replies: 9
    Last Post: June 9th 2008, 12:13 PM
  5. Area of triangle created from 3 planes
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 14th 2008, 06:27 AM

Search Tags


/mathhelpforum @mathhelpforum