При каких значениях параметра b уравнение 5(b+4)x^2-10x+b=0 имеет действительные корни...

Question 1

При каких значениях параметра b уравнение 5(b+4)x^2-10x+b=0 имеет действительные корни одного знака?

Question 2

Решите задачу:

$5(b+4)x^2-10x+b=0, \\ D_1=(-5)^2-5(b+4)b=25-5b^2-20b, \\ D \geq 0, \ -5b^2-20b+25 \geq 0, \\ b^2+4b-5 \leq 0, \\ b_1=-5, \ b_2=1, \\ (b+5)(b-1) \leq 0, \\ -5 \leq b \leq 1; \\ x=\frac{5\pm\sqrt{-5b^2-20b+25}}{5(b+4)}, \\ b+4 \neq 0, \ b \neq -4;$
$\left [ {{ \left \{ {{5-\sqrt{-5b^2-20b+25}\textless0,} \atop {5+\sqrt{-5b^2-20b+25}\textless0,}} \right. } \atop { \left \{ {{5-\sqrt{-5b^2-20b+25}\textgreater0,} \atop {5+\sqrt{-5b^2-20b+25}\textgreater0;}} \right. }} \right. \left [ {{ \left \{ {{\sqrt{-5b^2-20b+25}\textgreater5,} \atop {\sqrt{-5b^2-20b+25}\textless-5,}} \right. } \atop { \left \{ {{\sqrt{-5b^2-20b+25}\textless5,} \atop {\sqrt{-5b^2-20b+25}\textgreater-5;}} \right. }} \right.$
$\left [ {{ \left \{ {{-5b^2-20b+25\textgreater25,} \atop {b\in\varnothing,}} \right. } \atop { \left \{ {{-5b^2-20b+25\textless25,} \atop {-5b^2-20b+25\geq0;}} \right. }} \right. \left [ {{ b\in\varnothing,} \atop { \left \{ {{-5b^2-20b\textless0,} \atop {b^2+4b-5\leq 0;}} \right. }} \right. \left \{ {{(b+4)b\textgreater0,} \atop {(b+5)(b-1)\leq0;}} \right. \left \{ {{ \left[ {{b\textless-4,} \atop {b\textgreater0,}} \right. } \atop {-5 \leq b \leq 1;}} \right.$
$\left[ {{-5\leq b\textless-4,} \atop {0\ \textless \ b \leq 1.}} \right. \\ b\in[-5;-4)\cup(0;1]$