Question

Answer 1

\(\triangle=b^2-4ac=\left(-3\right)^2-4\cdot2\left(-6\right)=57>0\)

⇒ phương trình luôn có 2 nghiệm phân biệt

theo định lý vi -et ta có:

\(x_1+x_2=\frac32;x_1\cdot x_2=-\frac62=-3\)

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

\(=\left(\frac32\right)^2-4\cdot\left(-3\right)=\frac94+12=\frac{57}{4}\)

\(\left|x_1-x_2\right|=\sqrt{\left(x_1-x_2\right)^2}=\sqrt{\frac{57}{4}}\)