21/07/2009

Revival Post

Posted in math.GM at 13:27 by momar006

Alright, let’s revive this place !

How many real solutions are there to the equation

x^4 + 6x^2 + 2 = 4x(x^2 + 1).

2 Comments »

  1. ebw said,

    23/07/2009 at 00:57

    (x-1)^4 = x^4 + 6x^2 + 1 – 4x(x^2+1). Thus LHS>RHS for all real x.

    Reply

  2. David said,

    23/01/2010 at 14:03

    neat

    Reply

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