Xét phương trình \(2x^2-4x-1=0\)
\(\Delta=4^2-4.2.\left(-1\right)=24>0\) \(\Rightarrow\) PT có 2 nghiệm phân biệt \(x_1;x_2\)
Theo hệ thức Vi-et ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=-\dfrac{1}{2}\end{matrix}\right.\)
Đặt \(a=2x_1-x_2^2;b=2x_2-x_1^2\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2\left(x_1+x_2\right)-\left(x_1^2+x_2^2\right)\\ab=2x_1x_2-2x_1^3-2x_2^3+\left(x_1x_2\right)^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2\left(x_1+x_2\right)-\left(x_1+x_2\right)^2+2\left(x_1+x_2\right)\\ab=\left(x_1x_2\right)^2+2x_1x_2-2\left[\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\right]\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2.2-2^2+2.2=4\\ab=\left(-\dfrac{1}{2}\right)^2+2.\left(-\dfrac{1}{2}\right)-2\left(2^3-3.2.\left(-\dfrac{1}{2}\right)\right)=-\dfrac{91}{4}\end{matrix}\right.\)
Vậy PT bậc hai có 2 nghiệm \(a,b\) là \(X^2-4X-\dfrac{91}{4}=0\)
