Vì \(a\cdot c=1\cdot\left(-12\right)=-12<0\)
nên phương trình luôn có hai nghiệm phân biệt trái dấu
THeo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=-4\left(m-1\right);x_1x_2=\frac{c}{a}=-12\)
x2 là nghiệm của phương trình nên ta có:
\(x_2^2+4\left(m-1\right)\cdot x_2-12=0\)
=>\(x_2^2-4\cdot x_2+4+4mx_2-16=0\)
=>\(\left(x_2-2\right)^2=-4mx_2+16\)
=>\(\left(x_2-2\right)^2=-4\left(mx_2-4\right)=4\left(4-mx_2\right)\)
=>\(\sqrt{4\left(4-mx_2\right)}=\sqrt{\left(x_2-2\right)^2}=\left|x_2-2\right|\)
=>\(2\cdot\sqrt{4-mx_2}=\left|x_2-2\right|\)
\(4\cdot\left|x_1-2\right|\cdot\sqrt{4-mx_2}=\left(x_1+x_2-x_1x_2-8\right)^2\)
=>\(2\cdot\left|x_1-2\right|\cdot2\cdot\sqrt{4-mx_2}=\left(-4m+4+12-8\right)^2\)
=>\(2\left|x_1-2\right|\cdot\left|x_2-2\right|=\left(-4m+8\right)^2\)
=>\(2\cdot\left|x_1x_2-2\left(x_1+x_2\right)\right.+4\left|\right.=\left(-4m+8\right)^2\)
=>\(2\left|-12-2\left(-4m+4\right)+4\right|=\left(-4m+8\right)^2\)
=>2|-12+8m-8+4|=16(m-2)^2
=>2|8m-16|=16(m-2)^2
=>(m-2)^2=|m-2|
=>\(\left|m-2\right|\left(\left|m-2\right|-1\right)=0\)
TH1: |m-2|=0
=>m-2=0
=>m=2
TH2: |m-2|-1=0
=>|m-2|=1
=>\(\left[\begin{array}{l}m-2=1\\ m-2=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}m=3\left(nhận\right)\\ m=1\left(nhận\right)\end{array}\right.\)
