The quadratic equation is 3x^2+(1+2i)x+1-i=03x2+(1+2i)x+1−i=0
now its discriminant is (1+2i)^2-4*3*(1-i)(1+2i)2−4⋅3⋅(1−i)
= 1-4+4i-12+12i=-15+16i1−4+4i−12+12i=−15+16i
Hence using quadratic formula its roots are
x=(-(1+2i)+-sqrt(-15+16i))/6x=−(1+2i)±√−15+16i6
Let sqrt(-15+16i)=a+bi√−15+16i=a+bi, then squaring
-15+16i=a^2-b^2+2abi−15+16i=a2−b2+2abi
i.e. a^2-b^2=-15a2−b2=−15 and 2ab=162ab=16
and (a^2+b^2)^2=(a^2-b^2)^2+4a^2b^2=225+256=481(a2+b2)2=(a2−b2)2+4a2b2=225+256=481
i.e. a^2+b^2=sqrt481a2+b2=√481
and a^2=(sqrt481-15)/2a2=√481−152 and b^2=(sqrt481+15)/2b2=√481+152
i.e sqrt(-15+16i)=sqrt((sqrt481+15)/2)+isqrt((sqrt481+15)/2)√−15+16i=√√481+152+i√√481+152
and putting this we get
x=(-(1+2i)+-(sqrt((sqrt481+15)/2)+isqrt((sqrt481+15)/2)))/6x=−(1+2i)±(√√481+152+i√√481+152)6
= [-1+-sqrt((sqrt481+15)/2)]/6+(-2+-sqrt((sqrt481+15)/2))/6i−1±√√481+1526+−2±√√481+1526i
Note that if we choose ++ in real part, we choose ++ in imaginary part too and if we choose -− in real part, we choose -− in imaginary part too.
