TH1: m=0
Phương trình sẽ trở thành:
\(0x^2+2\left(0-4\right)x+0+7=0\)
=>-8x+7=0
=>\(x=\dfrac{7}{8}\)
=>Loại
TH2: \(m\ne0\)
\(\text{Δ}=\left[2\left(m-4\right)\right]^2-4m\left(m+7\right)\)
\(=4\left(m^2-8m+16\right)-4\left(m^2+7m\right)\)
\(=4\left(m^2-8m+16-m^2-7m\right)\)
\(=4\left(-15m+16\right)\)
Để phương trình có hai nghiệm thì Δ>=0
=>4(-15m+16)>=0
=>-15m+16>=0
=>-15m>=-16
=>\(m< =\dfrac{16}{15}\)
\(\left|x_1\right|=2\left|x_2\right|\)
=>\(\left[{}\begin{matrix}x_1=2x_2\\x_1=-2x_2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x_1-2x_2=0\\x_1+2x_2=0\end{matrix}\right.\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-2\left(m-4\right)}{m}=\dfrac{-2m+8}{m}\\x_1x_2=\dfrac{c}{a}=\dfrac{m+7}{m}\end{matrix}\right.\)
TH1: \(x_1=2x_2\)
\(x_1+x_2=\dfrac{-2m+8}{m}\)
=>\(3x_2=\dfrac{-2m+8}{m}\)
=>\(x_2=\dfrac{-2m+8}{3m}\)
=>\(x_1=\dfrac{2\left(-2m+8\right)}{3m}=\dfrac{-4m+16}{3m}\)
\(x_1\cdot x_2=\dfrac{m+7}{m}\)
=>\(\dfrac{\left(-2m+8\right)\left(-4m+16\right)}{9m^2}=\dfrac{m+7}{m}\)
=>\(\dfrac{8m^2-32m-32m+128}{9m^2}=\dfrac{9m\left(m+7\right)}{9m^2}\)
=>\(9m^2+63m=8m^2-64m+128\)
=>\(m^2+127m-128=0\)
=>(m+128)(m-1)=0
=>\(\left[{}\begin{matrix}m=-128\left(nhận\right)\\m=1\left(nhận\right)\end{matrix}\right.\)
TH2: \(x_1=-2x_2\)
\(x_1+x_2=\dfrac{-2m+8}{m}\)
=>\(-x_2=\dfrac{-2m+8}{m}\)
=>\(x_2=\dfrac{2m-8}{m}\)
=>\(x_1=\dfrac{-2\left(2m-8\right)}{m}=\dfrac{-4m+16}{m}\)
\(x_1x_2=\dfrac{m+7}{m}\)
=>\(\dfrac{\left(2m-8\right)\left(-4m+16\right)}{m^2}=\dfrac{m+7}{m}\)
=>\(\dfrac{-8m^2+32m+32m-128}{m^2}=\dfrac{m^2+7m}{m^2}\)
=>\(-8m^2+64m-128-m^2-7m=0\)
=>\(-9m^2+57m-128=0\)
\(\Leftrightarrow m\in\varnothing\)
