Question

Answer 1

Ta có: \(\Delta=\left(-6\right)^2-4\cdot1\cdot m\)

=36-4m

Để phương trình có hai nghiệm thì Δ>=0

=>36-4m>=0

=>4m<=36

=>m<=9

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

\(x_1^2-x_2^2=12\)

=>\(\left(x_1+x_2\right)\left(x_1-x_2\right)=12\)

=>\(6\left(x_1-x_2\right)=12\)

=>\(x_1-x_2=2\)

=>\(\left(x_1-x_2\right)^2=2^2=4\)

=>\(\left(x_1+x_2\right)^2-4x_1x_2=4\)

=>\(6^2-4m=4\)

=>4m=36-4=32

=>m=8(nhận)