\(x^2-2\left(2m-1\right)x+m^2-4m=0\)
\(\text{Δ }=\left[-2\left(2m-1\right)\right]^2-4\left(m^2-4m\right)\)
\(=4\left(2m-1\right)^2-4\left(m^2-4m\right)\)
\(=4\left(4m^2-4m+1\right)-4\left(m^2-4m\right)\)
\(=16m^2-16m+4-4m^2+16m\)
\(=12m^2+4>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left[-2\left(2m-1\right)\right]}{1}=2\left(2m-1\right)=4m-2\\x_1x_2=\dfrac{c}{a}=m^2-4m\end{matrix}\right.\)
\(x_1^2+x_2^2=10\)
=>\(\left(x_1+x_2\right)^2-2x_1x_2=10\)
=>\(\left(4m-2\right)^2-2\left(m^2-4m\right)=10\)
=>\(16m^2-16m+4-2m^2+8m-10=0\)
=>\(14m^2-8m-6=0\)
=>\(14m^2-14m+6m-6=0\)
=>\(\left(m-1\right)\left(14m+6\right)=0\)
=>\(\left[{}\begin{matrix}m=1\left(nhận\right)\\m=-\dfrac{3}{7}\left(nhận\right)\end{matrix}\right.\)
