Question

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

Answer 1

$5(b+4)x^2-10x+b=0, \ D_1=(-5)^2-5(b+4)b=25-5b^2-20b, \ D geq 0, -5b^2-20b+25 geq 0, \ b^2+4b-5 leq 0, \ b_1=-5, b_2=1, \ (b+5)(b-1) leq 0, \ -5 leq b leq 1; \ x=frac{5pmsqrt{-5b^2-20b+25}}{5(b+4)}, \ b+4 neq 0, b neq -4;$
$left [ {{ left { {{5-sqrt{-5b^2-20b+25}textless0,} atop {5+sqrt{-5b^2-20b+25}textless0,}} right. } atop { left { {{5-sqrt{-5b^2-20b+25}textgreater0,} atop {5+sqrt{-5b^2-20b+25}textgreater0;}} right. }} right. left [ {{ left { {{sqrt{-5b^2-20b+25}textgreater5,} atop {sqrt{-5b^2-20b+25}textless-5,}} right. } atop { left { {{sqrt{-5b^2-20b+25}textless5,} atop {sqrt{-5b^2-20b+25}textgreater-5;}} right. }} right.$
$left [ {{ left { {{-5b^2-20b+25textgreater25,} atop {binvarnothing,}} right. } atop { left { {{-5b^2-20b+25textless25,} atop {-5b^2-20b+25geq0;}} right. }} right. left [ {{ binvarnothing,} atop { left { {{-5b^2-20btextless0,} atop {b^2+4b-5leq 0;}} right. }} right. left { {{(b+4)btextgreater0,} atop {(b+5)(b-1)leq0;}} right. left { {{ left[ {{btextless-4,} atop {btextgreater0,}} right. } atop {-5 leq b leq 1;}} right.$
$left[ {{-5leq btextless-4,} atop {0 textless b leq 1.}} right. \ bin[-5;-4)cup(0;1]$