\(\Delta'=\left(m+1\right)^2-\left(m^2+2\right)=2m-1>0\Rightarrow m>\dfrac{1}{2}\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2+2\end{matrix}\right.\)
\(x_1^2+x_2^2=10\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=10\)
\(\Leftrightarrow4\left(m+1\right)^2-2\left(m^2+2\right)=10\)
\(\Leftrightarrow m^2+4m-5=0\Rightarrow\left[{}\begin{matrix}m=1\\m=-5< \dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)
Để pt có 2 nghiệm pb
\(\Delta'=\left(m+1\right)^2-m^2-2=2m+1-2=2m-1>0\Leftrightarrow m>\dfrac{1}{2}\)
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2+2\end{matrix}\right.\)
Ta có : \(\left(x_1+x_2\right)^2-2x_1x_2=10\)
Thay vào ta được \(4\left(m+1\right)^2-2\left(m^2+2\right)=10\Leftrightarrow4m^2+8m+4-2m^2-4=10\)
\(\Leftrightarrow2m^2+8m-10=0\)ta có : a + b + c = 0
vậy pt có nghiệm x = 1 ; x = -5 (ktmđk)
giúp mk vs