Để pt có 2 nghiệm phân biệt
⇒Δ'>0⇌(m+1)2-(m2+2)>0⇌m2+2m+1-m2-2>0⇌2m-1>0⇌m>\(\dfrac{1}{2}\)
⇒pt có 2 nghiệm phân biệt
\(x_1=m+1-\sqrt{2m-1}\); \(x_2=m+1+\sqrt{2m-1}\) \(\left(x_1\right)^2=\left(\left(m-1\right)-\sqrt{2m-1}\right)^2\)
=\(\left(m+1\right)^2-2\left(m+1\right)\sqrt{2m-1}+2m-1\)
TT:\(\left(x_2\right)^2\)=\(\left(m+1\right)^2+2\left(m+1\right)\sqrt{2m-1}+2m-1\)
⇒\(|\left(x_1\right)^2+\left(x_2\right)^2|\)
=\(|\left(m+1\right)^2-2\left(m+1\right)\sqrt{2m-1}+2m-1\)+
\(\left(m+1\right)^2+2\left(m+1\right)\sqrt{2m-1}+2m-1|\)
\(=|2\left(m+1\right)^2+2\left(2m-1\right)|\)
\(=|2\left(m^2+2m+1+2m-1\right)|\)
\(=2|m^2+4m|\)
Do \(m>\dfrac{1}{2}\)⇒m2+4m>0⇒\(=2|m^2+4m|\)=2m2+8m (1)
Lại có: \(|\left(x_1\right)^2+\left(x_2\right)^2|\)=16m2+64m (2)
Thay (1) vào (2) ta được:
2m2+8m=16m2+64m
⇌-14m2-56m=0
⇌-14m(m-4)=0
⇌m(m-4)=0
⇌\(\left[{}\begin{matrix}m=0\\m-4=0\end{matrix}\right.\) ⇌\(\left[{}\begin{matrix}m=0\\m=4\end{matrix}\right.\)
Do \(m>\dfrac{1}{2}\)⇒m=4
Vậy với m=4 thì pt có 2 nghiệm phân biệt thỏa mãn:
\(|\left(x_1\right)^2+\left(x_2\right)^2|\)=16m2+64m
