\(\Delta'=\left[-2\left(m+2\right)\right]^2-2\cdot\left(2m^2+1\right)\)
\(=4\left(m^2+4m+4\right)-4m^2-2\\ =4m^2+16m+16-4m^2-2\\ =16m+14\)
Để pt có nghiệm \(\Leftrightarrow\Delta'\ge0\Leftrightarrow16m+14\ge0\)
\(\Leftrightarrow16m\ge-14\\ m\ge\dfrac{-7}{8}\)
Với \(m\ge\dfrac{-7}{8}\) theo vi-ét ta có :
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+2\right)\\x_1x_2=\dfrac{2m^2+1}{2}\end{matrix}\right.\)
Ta có : \(x^2_1+x_2^2=\dfrac{15}{2}\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=\dfrac{15}{2}\)
\(\Leftrightarrow\left[2\left(m+2\right)\right]^2-2\cdot\dfrac{2m^2+1}{2}=\dfrac{15}{2}\)
\(\Leftrightarrow4\left(m^2+4m+4\right)-2m^2-1=\dfrac{15}{2}\)
\(\Leftrightarrow4m^2+16m+16-2m^2-1-\dfrac{15}{2}=0\)
\(\Leftrightarrow2m^2+16m+\dfrac{15}{2}=0\)
\(\Leftrightarrow4m^2+32m+15=0\)
\(\Delta_m=32^2-4\cdot4\cdot15=1024-240=784>0\)
\(\Rightarrow\sqrt{\Delta}=\sqrt{784}=28\)
Vì \(\Delta>0\) nên pt có 2 nghiệm phân biệt
\(\Rightarrow m_1=\dfrac{-32+28}{2\cdot4}=-\dfrac{1}{2}\) (m TM \(m\ge\dfrac{-7}{8}\))
\(m_2=\dfrac{-32-28}{2\cdot4}=-\dfrac{15}{2}\) (m KTM \(m\ge\dfrac{-7}{8}\))
Vậy \(m=-\dfrac{1}{2}\) để 2 nghiệm \(x_1^2;x^2_2\) thỏa mãn \(x_1^2+x^2_2=\dfrac{15}{2}\)
Vậy ...........................
