a: Thay m=-3 vào (1), ta được:
\(x^2-\left(-3-2\right)x-6=0\)
=>\(x^2+5x-6=0\)
=>(x+6)(x-1)=0
=>\(\left[{}\begin{matrix}x=-6\\x=1\end{matrix}\right.\)
b: \(a\cdot c=1\cdot\left(-6\right)=-6< 0\)
=>phương trình (1) luôn có hai nghiệm phân biệt
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=m-2\\x_1x_2=\dfrac{c}{a}=-6\end{matrix}\right.\)
\(x_2^2+x_1x_2+\left(m-2\right)\cdot x_1=16\)
=>\(x_2^2+x_1x_2+x_1\left(x_1+x_2\right)=16\)
=>\(x_1^2+2x_1x_2+x_2^2=16\)
=>\(\left(x_1+x_2\right)^2=16\)
=>\(\left(m-2\right)^2=16\)
=>\(\left[{}\begin{matrix}m-2=4\\m-2=-4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}m=6\\m=-2\end{matrix}\right.\)
