Question

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

Tìm m để \(B=\left(x_1-x_2\right)^2\) đạt GTNN. 

Answer 1

PT có 2 nghiệm \(\Leftrightarrow\Delta=\left(4m+1\right)^2-8\left(m-4\right)\ge0\)

\(\Leftrightarrow16m^2+33\ge0\left(\text{luôn đúng}\right)\)

Áp dụng Viét: \(\left\{{}\begin{matrix}x_1+x_2=4m+1\\x_1x_2=-2\left(m-4\right)\end{matrix}\right.\)

\(B=\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2=\left(4m+1\right)^2+8\left(m-4\right)\\ B=16m^2+16m-31=4\left(4m^2+4m+1\right)-35=4\left(2m+1\right)^2-35\ge-35\)

Vậy \(B_{min}=-35\Leftrightarrow m=-\dfrac{1}{2}\)