Question

Answer 1

\(\Delta=\left(-6\right)^2-4\left(m+3\right)\)

=36-4m-12

=-4m+24

Để phương trình có hai nghiệm phân biệt thì Δ>0

=>-4m+24>0

=>-4m>-24

=>m<6

Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=6;x_1x_2=\frac{c}{a}=m+3\)

\(x_1+x_2=6\)

=>\(x_2=6-x_1\)

\(x_2=x_1^2\)

=>\(x_1^2=6-x_1\)

=>\(x_1^2+x_1-6=0\)

=>\(\left(x_1+3\right)\left(x_1-2\right)=0\)

=>\(\left[\begin{array}{l}x_1=-3\\ x_1=2\end{array}\right.\)

TH1: \(x_1=-3\)

=>\(x_2=6-\left(-3\right)=6+3=9\)

\(x_1\cdot x_2=m+3\)

=>m+3=-27

=>m=-30(nhận)

TH2: \(x_1=2\)

=>\(x_1=6-2=4\)

\(x_1x_2=m+3\)

=>\(m+3=2\cdot4=8\)

=>m=5(nhận)