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.
