\(\text{Δ}=\left(-2\right)^2-4\cdot1\cdot\left(m-3\right)\)
\(=4-4m+12=16-4m\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>-4m+16>0
=>-4m>-16
=>m<4
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left(-2\right)}{1}=2\\x_1x_2=\dfrac{c}{a}=\dfrac{m-3}{1}=m-3\end{matrix}\right.\)
\(x_1^2-2x_2+x_1x_2=-12\)
=>\(x_1^2-x_2\left(x_1+x_2\right)+x_1x_2=-12\)
=>\(x_1^2-x_2x_1-x_2^2+x_1x_2=-12\)
=>\(x_1^2-x_2^2=-12\)
=>\(\left(x_1-x_2\right)\left(x_1+x_2\right)=-12\)
=>\(x_1-x_2=-6\)
=>\(\left(x_1-x_2\right)^2=36\)
=>\(\left(x_1+x_2\right)^2-4x_1x_2=36\)
=>\(2^2-4\cdot\left(m-3\right)=36\)
=>4(m-3)=4-36=-32
=>m-3=-8
=>m=-5(nhận)
\(\Delta'=1-\left(m-3\right)=4-m\)
Pt có 2 nghiệm pb khi \(4-m>0\Rightarrow m< 4\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m-3\end{matrix}\right.\)
Do \(x_1\) là nghiệm nên:
\(x_1^2-2x_1+m-3=0\Rightarrow x_1^2=2x_1-m+3\)
Thay vào:
\(2x_1-m+3-2x_2+x_1x_2=-12\)
\(\Leftrightarrow2\left(x_1-x_2\right)-m+3+m-3=-12\)
\(\Leftrightarrow x_1-x_2=-6\)
Kết hợp với Viet \(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=2\\x_1-x_2=-6\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=-2\\x_2=4\end{matrix}\right.\)
Thế vào \(x_1x_2=m-3\Rightarrow m-3=-8\)
\(\Rightarrow m=-5\left(tm\right)\)
