Question 1160751


 {{{(z+1)^2 = 2z^2}}}

 solve for {{{z}}}:

{{{z^2+2z+1 = 2z^2}}}

{{{ 2z^2-z^2-2z-1=0}}}

{{{ z^2-2z-1=0}}}.....use quadratic formula

{{{z=(-b+-sqrt(b^2-4ac))/2a}}}


{{{z=(-(-2)+-sqrt((-2)^2-4*1(-1)))/(2*1)}}}


{{{z=(2+-sqrt(4+4))/2}}}

{{{z=(2+-sqrt(2*4))/2}}}

{{{z=(2+-2sqrt(2))/2}}}

{{{z=(1+-sqrt(2))}}}

solutions:

{{{z=1+sqrt(2)}}}

{{{z=1-sqrt(2)}}}