Question

Помогите пожалуйста=)=)=)=)=)

Answer 1

$ax^2 +bx + c=0; \ \ x_{1,2}=\frac{b^2 \pm \sqrt{b^2 - 4 \cdot a \cdot c}}{2 \cdot a} \\ \\ 3x^2 +7x-11=0; \ \ \ x_{1,2}=\frac{-7 \pm \sqrt{49 +132 }}{6}=\frac{-7 \pm \sqrt{181}}{6} \\ \\ 2 \cdot (\frac{-7 +\sqrt{181}}{6}) \cdot (\frac{-7 - \sqrt{181}}{6}) - (\frac{-7 + \sqrt{181}}{6}) - (\frac{-7 - \sqrt{181}}{6})=\\ \\ = 2 \cdot (\frac{49 -181}{36}) +\frac{7 -\sqrt{181}+7 + \sqrt{181}}{6}=2 \cdot (\frac{-132}{36}) + \frac{14}{6} =-\frac{132}{18}+ \frac{14}{6} =\\\\= \frac{-132 +42}{18}=\frac{-90}{18}=-5$

Answer 2

$3x^2+7x-11=0\; \; \Rightarrow \; \; \left \{ {{x_1\cdot x_2=-\frac{11}{3}} \atop {x_1+x_2=-\frac{7}{3}}} \right. \; \; (teor.\; Vieta)\\\\2x_1\cdot x_2-x_1-x_2=2x_1\cdot x_2-(x_1+x_2)=2\cdot (-\frac{11}{3})-(-\frac{7}{3})=\\\\=\frac{-22}{3}+\frac{7}{3}=\frac{-15}{3}=-5$