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

Ответы 1

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