Question

решите 9 номер пожалуйста отношение корней уравнения ..... равны 6

Answer 1

$32x^{2}+bx+75=0\\\\x^{2}+\frac{b}{32}x+\frac{75}{32} =0\\\\\frac{x_{1} }{x_{2} }=6 \ \Rightarrow \ x_{1}=6x_{2}\\\\x_{1} *x_{2} =\frac{75}{32}\\\\6x_{2} *x_{2}=\frac{75}{32}\\\\6x_{2}^{2}=\frac{75}{32}\\\\x_{2}^{2}=\frac{25}{64}\\\\x'_{2}=-\frac{5}{8}\\\\x''_{2}=\frac{5}{8} \\\\x'_{1}=6*(-\frac{5}{8} )=-\frac{15}{4}\\\\x''_{1}=6*\frac{5}{8}=\frac{15}{4} \\\\1)-\frac{b}{32} =x'_{1}+x'_{2}\\\\-\frac{b}{32}=-\frac{15}{4}-\frac{5}{8}$

$-\frac{b}{32}=-\frac{35}{8}\\\\-\frac{b}{32}=-\frac{140}{32}\\\\b_{1}=140\\\\2)-\frac{b}{32}=x'_{2}+x''_{2} }\\\\-\frac{b}{32} =\frac{5}{8}+\frac{15}{4}\\\\-\frac{b}{32} =\frac{35}{8}\\\\-\frac{b}{32} =\frac{140}{32}\\\\b_{2} =-140\\\\\boxed{-140 \ ; \ 140}$