Question

x^2+2x+q=0
Отношение корней=6

Answer 1

$x^2+2x+q=0\; ,\; \; \frac{x_1}{x_2}=6\; \; \to \; \; x_1=6x_2\\\\Teor.\; Vieta\; :\; \; \left \{ {x_1\cdot x_2=q} \atop {x_1+x_2=-2}} \right. \; \left \{ {{6x_2^2=q} \atop {7x_2=-2}} \right. \; \left \{ {{q=6\cdot \frac{4}{49}=\frac{24}{49}} \atop {x_2=-\frac{2}{7}}} \right. \\\\x_1=6\cdot \frac{-2}{7}=-\frac{12}{7}$