Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) . Khi đó hệ phương trình trở thành :
\(\left\{{}\begin{matrix}a+b=-1\\a^2-a-2b=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-a-1\left(1\right)\\a^2+a-20=0\left(2\right)\end{matrix}\right.\)
Xét phương trình (2) : \(a^2+a-20=0\)
\(\Delta=1+80=81>0\)
\(\Rightarrow\left\{{}\begin{matrix}a_1=\frac{-1+9}{2}=4\\a_1=\frac{-1-9}{2}=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b_1=-5\\b_2=4\end{matrix}\right.\)
Với \(\left(a_1;b_1\right)=\left(4;-5\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=4\\xy=-5\end{matrix}\right.\Rightarrow x^2-4x-5=0\)
\(\Delta=16+20=36>0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\frac{4+6}{2}=5\\x_2=\frac{4-6}{2}=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y_1=-1\\y_2=5\end{matrix}\right.\)
Với \(\left(a_2;b_2\right)=\left(-5;4\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=-5\\xy=4\end{matrix}\right.\Rightarrow x^2+5x+4=0\)
\(\Delta=25-16=9>0\)
\(\Rightarrow\left\{{}\begin{matrix}x_3=\frac{-5+3}{2}=-1\\x_4=\frac{-5-3}{2}=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y_3=-4\\y_4=-1\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}\left(x_1;y_1\right)=\left(5;-1\right)\\\left(x_1;y_2\right)=\left(-1;5\right)\\\left(x_3;y_3\right)=\left(-1;-4\right)\\\left(x_4;y_4\right)=\left(-4;-1\right)\end{matrix}\right.\)
