Question

Answer 1

Câu 36: \(\text{Δ}=\left(m+2\right)^2-4\cdot\left(-2\right)\left(-m-4\right)\)

\(=m^2+4m+4+8\left(-m-4\right)\)

\(=m^2+4m+4-8m-32=m^2-4m-28\)

Để phương trình có nghiệm thì Δ>=0

=>\(m^2-4m-28>=0\)

=>\(m^2-4m+4-32>=0\)

=>\(\left(m-2\right)^2>=32\)

=>\(\left[{}\begin{matrix}m-2>=4\sqrt{2}\\m-2< =-4\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m>=4\sqrt{2}+2\\m< =-4\sqrt{2}+2\end{matrix}\right.\)