\(\Delta=\left[-2\left(m-1\right)\right]^2-4\cdot1\cdot\left(m^2-3\right)\)
\(=4\left(m^2-2m+1\right)-4\left(m^2-3\right)\)
\(=4m^2-8m+4-4m^2+12=-8m+16\)
Để phương trình có hai nghiệm thì \(\Delta>=0\)
=>-8m+16>=0
=>-8m>=-16
=>m<=2
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m-1\right)=2m-2\\x_1x_2=\dfrac{c}{a}=m^2-3\end{matrix}\right.\)
\(\left|x_1-x_2\right|=\sqrt{22-x_1x_2}\)
=>\(\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=\sqrt{22-x_1x_2}\)
=>\(\sqrt{\left(2m-2\right)^2-4\left(m^2-3\right)}=\sqrt{22-\left(m^2-3\right)}\)
=>\(4m^2-8m+4-4m^2+12=22-m^2+3\)
=>\(-8m+16=-m^2+25\)
=>\(-8m+16+m^2-25=0\)
=>\(m^2-8m-9=0\)
=>(m-9)(m+1)=0
=>\(\left[{}\begin{matrix}m=9\left(loại\right)\\m=-1\left(nhận\right)\end{matrix}\right.\)
