The complex roots of a quadratic always come in pairs. If one root is #(1+2i)# then the other root must be #(1-2i)#. This must be true if the quadratic has real coefficients, since we know that:
If #alpha# and #beta# are the roots of a quadratic, then:
#alpha + beta = (-b)/(2a)# and #alpha *beta = c/a#
This is only possible with complex conjugates.
If:
#x = (1+2i)# then #x-(1+2i)=0 color(white)(*) color(blue)([1])#
If:
#x=(1-2i)# then #x-(1-2i)=0 color(white)(*)color(blue)([2])#
Multiply #color(blue)([1]) andcolor(blue)([2])#
#{x-(1+2i)}*{x-(1-2i)}=x^2-(1+2i)*x-(1-2i)*x+(1+2i)(1-2i)#
#-> = x^2 -x-2ix-x+2ix+1-4i^2#
#->=x^2-2x+1 -(4*-1)= color(blue)(x^2-2x+5)#
Check with quadratic formula:
#x= (-(-2)+-sqrt(4-20))/2=>x= 1+2i , x=1-2i#
