As the quadratic equation has its solutions #2/3+-(sqrt2i)/3#, two complex conjugate numbers, it should have real coefficients. Thereare two ways in which this can be worked out,
Method I #-# In this method as #2/3+-(sqrt2i)/3# are two solutions, equation is
#(x-(2/3+(sqrt2i)/3))(x-(2/3-(sqrt2i)/3))=0#
or #(x-2/3-(sqrt2i)/3)(x-2/3+(sqrt2i)/3)=0#
i.e. #(x-2/3)^2-((sqrt2i)/3)^2=0#
i.e. #x^2-4/3x+4/9+2/9=0#
or #x^2-4/3x+6/9=0#
or #3x^2-4x+2=0#
Method II #-# In this method, we use the fact that if two roots are #alpha# and #beta#, equation is
#x^2-(alpha+beta)x+alphabeta=0#
As sum of roots is #4/3# and product of roots is #(2/3)^2+2/9=6/9=2/3# (note that product of two complex conjugate numbers #a+-bi# is #a^2+b^2#), equation is
#x^2-4/3x+2/3=0# or #3x^2-4x+2=0#
Note - The FOIL rule converts a product of two binomials into a sum of four monomials. Reverse foil method is the process of changing the form of an expression from one containing terms to factors. Here as factors are known - #alpha=2/3-(sqrt2i)/3# and #beta=2/3-(sqrt2i)/3#, we can use FOIL rule and not reverse FOIL. Using FOIL rule we have #(x-alpha)(x-beta)=x*x+x*(-beta)+x*(-alpha)+(-alpha)(-beta)#
= #x^2-alphax-betax+alphabeta#
= #x^2-(alpha+beta)x+alphabeta#,
which is used in method II.
