Question 49269
if the roots are (4+i)and (4-i), then we can say that:


x=(4+i) or x=(4-i)
hence
x-(4+i) = 0 or x-(4-i)=0


Working backwards we then say 
[x-(4+i)][x-(4-i)] = 0
(x-4-i)(x-4+i) = 0
{{{ x^2-4x+ix-4x+16-4i-ix+4i-i^2 = 0 }}}
{{{ x^2-8x+ix+16-4i-ix+4i-i^2 = 0 }}}
{{{ x^2-8x+16-4i+4i-i^2 = 0 }}}
{{{ x^2-8x+16-i^2 = 0 }}}
{{{ x^2-8x+16-(-1) = 0 }}}
{{{ x^2-8x+16+1 = 0 }}}
{{{ x^2-8x+17 = 0 }}}


there you go.


jon.