Odpowiedź:
[tex]\huge\boxed{m\in\left(1;\ 10\right>}[/tex]
Szczegółowe wyjaśnienie:
[tex]\left\{\begin{array}{ccc}a\neq0&(1)\\\Delta\geq0&(2)\\x_1\cdot x_2>0&(3)\\x_1+x_2<0&(4)\end{array}\right[/tex]
[tex]x^2+6x+m-1=0\\\\a=1,\ b=6,\ c=m-1[/tex]
[tex](1)\\a=1\neq0\Rightarrow\huge\boxed{m\in\mathbb{R}}[/tex]
[tex](2)\\\Delta=6^2-4\cdot1\cdot(m-1)=36-4m+4=-4m+40\\\\\Delta\geq0\iff-4m+40\geq0\qquad|-40\\\\-4m\geq-40\qquad|:(-4)\\\\\huge\boxed{m\leq10\to m\in\left(-\infty,\ 10\right>}[/tex]
W (3) i (4) mamy zastosowanie wzorów Viete'a:
[tex]x_1\cdot x_2=\dfrac{c}{a}\\\\x_1+x_2=\dfrac{-b}{a}[/tex]
[tex](3)\\x_1\cdot x_2=\dfrac{m-1}{1}>0\\\\m-1>0\qquad|+1\\\\\huge\boxed{m>1\to m\in(1,\ \infty)}[/tex]
[tex](4)\\\\x_1+x_2=\dfrac{-6}{1}<0\\\\-6<0\qquad\bold{PRAWDA}\\\\\huge\boxed{m\in\mathbb{R}}[/tex]
Z (1), (2), (3) i (4) mamy:
[tex]\huge\boxed{m\in\left(1;\ 10\right>}[/tex]
