Question

Не вычисляя корней уравнения 3х²+8х-1=0, найдите: а) x1+x2 б)x1*x2 в)1/x1 + 1/x2 и все че там еще сеть

Answer 1

$3x^{2}+5x-1=0\\\\1)x_{1} +x_{2}=-\frac{5}{3}\\\\2)x_{1} *x_{2}=-\frac{1}{3}\\\\3)\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1}*x_{2}}=\frac{-\frac{5}{3}}{-\frac{1}{3}}=\frac{5}{3}*3=5\\\\ 4)x_{1}^{2}+x_{2}^{2}=(x_{1}+x_{2})^{2}-2x_{1}x_{2}=(-\frac{5}{3})^{2}-2*(-\frac{1}{3})=\frac{25}{9} +\frac{2}{3}=\\\\=\frac{25}{9} +\frac{6}{9}=\frac{31}{9}=3\frac{4}{9}$

$5)(x_{1}-x_{2})^{2}=x_{1}^{2}-2x_{1}x_{2}+x_{2}^{2} =(x_{1}+x_{2})^{2}-2x_{1}x_{2}-2x_{1}x_{2}=\\\\=(x_{1}+x_{2})^{2}-4x_{1}x_{2}=(-\frac{5}{3})^{2}-4*(-\frac{1}{3})=\frac{25}{9}+\frac{4}{3}=\frac{25}{9}+\frac{12}{9}=\frac{37}{9}=4\frac{1}{9}\\\\\frac{x_{1} }{x_{2}}+\frac{x_{2}}{x_{1}}=\frac{x_{1}^{2}+x_{2}^{2}}{x_{1}x_{2}}=\frac{(x_{1}+x_{2})^{2}-2x_{1}x_{2}}{x_{1}x_{2}}=3\frac{4}{9}:(-\frac{1}{3})=-\frac{31}{9}*3=-\frac{31}{3}=-10\frac{1}{3}$