Question

\(x^2-2\left(m+1\right)x+m^2+2=0\left(1\right)\)

tìm m để (1) có 2 nghiệm phân biệt `x_1 ,x_2` thỏa mãn \(x_1^2+2\left(m+1\right)x_2=12m+2\)

Answer 1

\(\Delta'=\left(m+1\right)^2-\left(m^2+2\right)=2m-1\)

Pt có 2 nghiệm pb khi \(2m-1>0\Rightarrow m>\dfrac{1}{2}\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2+2\end{matrix}\right.\)

Do \(x_1\) là nghiệm pt nên:

\(x_1^2-2\left(m+1\right)x_1+m^2+2=0\Rightarrow x_1^2=2\left(m+1\right)x_1-m^2-2\)

Từ đó ta có:

\(x_1^2+2\left(m+1\right)x_2=12m+2\)

\(\Leftrightarrow2\left(m+1\right)x_1-m^2-2+2\left(m+1\right)x_2=12m+2\)

\(\Leftrightarrow2\left(m+1\right)\left(x_1+x_2\right)-m^2-12m-4=0\)

\(\Leftrightarrow4\left(m+1\right)^2-m^2-12m-4=0\)

\(\Leftrightarrow3m^2-4m=0\Rightarrow\left[{}\begin{matrix}m=0< \dfrac{1}{2}\left(loại\right)\\m=\dfrac{4}{3}\end{matrix}\right.\)