\(\Delta=4\left(m-1\right)^2-4\left(-3-m\right)=4m^2-4m+16=\left(2m-1\right)^2+15>0\forall m\)
\(\Rightarrow\) Pt có 2 nghiệm phân biệt với mọi m
Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m-2\\x_1x_2=-3-m\end{matrix}\right.\)
Theo đề ra, ta có: \(x_1^2+x_2^2=10\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=10\)
\(\Rightarrow\left(2m-2\right)^2-2\left(-3-m\right)=10\)
\(\Leftrightarrow4m^2-8m+4+6+2m=10\)
\(\Leftrightarrow4m^2-6m+10=10\)
\(\Leftrightarrow2m\left(2m-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=0\\2m-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=0\\m=\dfrac{3}{2}\end{matrix}\right.\)
\(\Delta'=\left[-\left(m-1\right)\right]^2-1\cdot\left(-3-m\right)=m^2-2m+1+3+m=m^2-m+4=\left(m-\dfrac{1}{2}\right)^2+\dfrac{15}{4}>0\forall m\)\(\Rightarrow\) PT luôn có 2 nghiệm phân biệt
Theo Vi-ét ta có:\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1\cdot x_2=-3-m\end{matrix}\right.\)
Theo đề bài ta có:\(x_1^2+x_2^2=10\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=10\Leftrightarrow\left[2\left(m-1\right)\right]^2-2\left(-3-m\right)=10\)
\(\Leftrightarrow4m^2-8m+4+6+2m-10=0\)
\(\Leftrightarrow4m^2-6m=0\)
\(\Leftrightarrow m\left(4m-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=0\\m=\dfrac{3}{2}\end{matrix}\right.\)
\(\Delta=\left[-2\left(m-1\right)\right]^2-4\cdot1\cdot\left(-m-3\right)\)
\(=4m^2-8m+4+4m+12=4m^2-4m+16\)
\(=4m^2-4m+1+15=\left(2m-1\right)^2+15>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m-1\right)\\x_1x_2=\dfrac{c}{a}=-m-3\end{matrix}\right.\)
\(x_1^2+x_2^2=10\)
=>\(\left(x_1+x_2\right)^2-2x_1x_2=10\)
=>\(\left(2m-2\right)^2-2\left(-m-3\right)=10\)
=>\(4m^2-8m+4+2m+6-10=0\)
=>\(4m^2-6m=0\)
=>\(2m^2-3m=0\)
=>m(2m-3)=0
=>\(\left[{}\begin{matrix}m=0\\m=\dfrac{3}{2}\end{matrix}\right.\)
