Question

Giải phương trình \(z^{2} +(1-i)z-18+13i=0.\)

\(A. z_{1} =4-i,z_{2} =-5+2i.  \)

\(B. z_{1} =4-i,z_{2} =-5-2i. \)

\(C. z_{1} =4-i,z_{2} =5-2i. \)

\(D. z_{1} =4+i,z_{2} =5+2i.\)

Answer 1

Ta chọn câu A
\(z^{2} +(1-i)z-18+13i=0. \)
Ta có: \(\Delta =(1-i)^{2} -4(-18+13i)=72-54i=(9-3i)^{2} .\)
\(\Rightarrow \left[\begin{array}{l} {z_{1} =\frac{-b+\sqrt{\Delta } }{2a} =4-i} \\ {z_{2} =\frac{-b-\sqrt{\Delta } }{2a} =-5+2i} \end{array}\right. . \)