Question

Тема: комплексные числа

Answer 1

$(1+i)x^2+(4-2i)x+(-1-7i)=0\ D/4=(2-i)^2+(1+i)(1+7i)=-3+4i=1+4i-4=(1+2i)^2\ x=dfrac{(-2+i)pm(1+2i)}{1+i}=dfrac12(1-i)((-2+i)pm(1+2i))=\ =dfrac12[(-1+3i)pm(3+i)]\ xin{-2+i, 1+2i}$