Question

Fie ecuatia x²+nx+n-1=0 ,n∈R si radaciniile ecuatiei , x₁ si x₂.Sa se afle multimea valorilor lui n pentru care x₁

Answer 1

 

$\displaystyle\bf\\x^2+nx+n-1=0\\\\x_{12}=\frac{-n\pm\sqrt{n^2-4(n-1)}}{2}\\\\x_{12}=\frac{-n\pm\sqrt{(n-2)^2}}{2}\\\\x_{12}=\frac{-n\pm(n-2)}{2}\\\\x_1=\frac{-n+n-2}{2}=\frac{-2}{2}=\boxed{\bf-1}\\\\x_2=\frac{-n-(n-2)}{2}=\frac{-n-n+2}{2}=\frac{-2n+2}{2}=\boxed{\bf-n+1}\\\\\\Punem~conditia~din~problema:\\\\x_1<n<x_2\\\\-1<n<-n+1\\\\-1<n \implies~\boxed{\bf~n>-1}\\\\n<-n+1\\\\n+n<1\\\\2n<1\\\\\boxed{\bf~n<\frac{1}{2}}\\\\Solutia~problemei:\\\\\boxed{\bf~n\in\left(-1,~\frac{1}{2}\right)}$