Ta có: \(x^2-4x+m+1=0\)
a=1; b=-4; c=m+1
\(\Delta=b^2-4ac\)
\(=\left(-4\right)^2-4\cdot1\cdot\left(m+1\right)\)
\(=16-4m-4\)
\(=-4m+12\)
Để phương trình (1) có hai nghiệm x1,x2 thì \(\Delta\ge0\)
\(\Leftrightarrow-4m+12\ge0\)
\(\Leftrightarrow-4m\ge-12\)
hay \(m\le3\)
Áp dụng hệ thức Vi-et, ta được:
\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left(-4\right)}{1}=4\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{m+1}{1}=m+1\end{matrix}\right.\)
Ta có: \(\left|x_1-x_2\right|=3m-4\)
\(\Leftrightarrow\sqrt{\left(x_1-x_2\right)^2}=3m-4\)
\(\Leftrightarrow\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=3m-4\)
\(\Leftrightarrow\sqrt{4^2-4\left(m+1\right)}=3m-4\)
\(\Leftrightarrow\sqrt{16-4m-4}=3m-4\)
\(\Leftrightarrow\sqrt{-4m+12}=3m-4\)
\(\Leftrightarrow-4m+12=\left(3m-4\right)^2\)
\(\Leftrightarrow-4m+12=9m^2-24m+16\)
\(\Leftrightarrow9m^2-24m+16+4m-12=0\)
\(\Leftrightarrow9m^2-20m+4=0\)(2)
a=9; b=-20; c=4
\(\Delta=b^2-4ac\)
\(=\left(-20\right)^2-4\cdot9\cdot4=400-144=256\)
Vì \(\Delta>0\) nên phương trình (2) có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}m_1=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{20-16}{18}=\dfrac{4}{18}=\dfrac{2}{9}\left(nhận\right)\\m_2=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{20+16}{18}=\dfrac{36}{18}=2\left(nhận\right)\end{matrix}\right.\)
Vậy: \(m\in\left\{\dfrac{2}{9};2\right\}\)
