Question

Answer 1

a: \(\text{Δ}=\left[-\left(m+1\right)\right]^2-4\cdot1\cdot\left(2m-3\right)\)

\(=\left(m+1\right)^2-4\left(2m-3\right)\)

\(=m^2+2m+1-8m+12=m^2-6m+13\)

\(=m^2-6m+9+4=\left(m-3\right)^2+4>0\forall m\)

=>PHương trình luôn có 2 nghiệm phân biệt

b: Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=m+1\\x_1x_2=\dfrac{c}{a}=2m-3\end{matrix}\right.\)

\(\left(x_1^2+x_2^2\right)+x_1x_2=8\)

=>\(\left(x_1+x_2\right)^2-x_1x_2=8\)

=>\(\left(m+1\right)^2-\left(2m-3\right)-8=0\)

=>\(m^2+2m+1-2m+3-8=0\)

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

=>\(m^2=4\)

=>\(\left[{}\begin{matrix}m=2\\m=-2\end{matrix}\right.\)