Question 1116475
{{{ (2x-1)^4 + (2x-1)^2 - 6 = 0 }}}

Let {{{ u = (2x-1)^2 }}}

{{{ u^2 + u - 6 = 0 }}}
{{{ (u-2)(u+3) = 0 }}}

u=2  and u=-3 are solutions, consider each in turn:<br>

u=2:   {{{ (2x-1)^2 = 2 }}}
           {{{ 4x^2 - 4x + 1 = 2 }}}
           {{{ 4x^2 - 4x - 1  = 0 }}}
           {{{ x^2 - x - (1/4) = 0 }}}
           {{{ x^2 - x   = (1/4) }}}
           {{{ x^2 - x + (1/4)  = (1/4) + (1/4) }}} 
           {{{   (x - 1/2)^2  = 1/2 }}}
           {{{  x - 1/2  }}} = +/- {{{ 1/sqrt(2)  }}}
           {{{  x }}} = +/- {{{ sqrt(2)/2 + (1/2) }}} <br>

Two of the four solutions are real solutions:
            {{{ highlight(  x= (1-sqrt(2))/2 ) }}}   and  {{{highlight( x=(1+sqrt(2))/2 ) }}} <br>

Now for the complex solutions:
u=-3:  {{{ (2x-1)^2 = -3 }}}
            {{{  4x^2 - 4x + 1 = -3 }}}
            {{{  4x^2 - 4x  = -4 }}}
            {{{  x^2 - x = -1 }}}   
             {{{ x^2 -x + 1/4 = -3/4 }}}
            {{{ (x-1/2)^2 = -3/4 }}}
           {{{   x-1/2 }}} =  +/- {{{ sqrt(3)i/2 }}}
     
Which has complex solutions:
            {{{  highlight( x= 1/2 - i*(sqrt(3)/2) ) }}}  and  {{{ highlight( x=1/2 +i*(sqrt(3)/2)) }}} <br>

The four roots together are all of the solutions to the 4th order equation.