Question

(x+3)(x+1)(x+5)(x+7)=9 - найти сумму корней уравнения

Answer 1

$(x+3)(x+1)(x+5)(x+7)=9\\\\(x+3)(x+5)=x^2+8x+15\\\\(x+1)(x+7)=x^2+8x+7\\\\t=x^2+8x+7\; \; \to \; \; x^2+8x+15=t+8\\\\t(t+8)=9\\\\t^2+8t-9=0\\\\t_1=1,t_2=-9\; (teor.\; Vieta)\\\\a)\; x^2+8x+7=1\\\\x^2+8x+6=0$

$D=40,x_1=\frac{-8-\sqrt{40}}{2}=-4-\sqrt{10}\\\\x_2=-4+\sqrt{10}\\\\b)\; x^2+8x+7=-9\\\\x^2+8x+16=0\\\\(x+4)^2=0\\\\x=-4\\\\Otvet:\; x_1=-4-\sqrt{10},x_2=-4+\sqrt{10},x_3=-4$