Question

Se considera funcția (m+2)x²+2mx+m-1 cu soluțiile x1<1,x2<1.Sa se determine m pentru fiecare soluție.

Answer 1

[tex]\triangle \geq 0=\ \textgreater \ m\in(-\infty,2]\\ x_1\ \textless \ 1\\ x_2\ \textless \ 1\\ x_1+x_2\ \textless \ 2=\ \textgreater \ -\frac{2m}{m+1} \ \textless \ 2=\ \textgreater \ m\in(-\infty,-2)U(-1,+\infty)\\ (x_1-1)(x_2-1)\ \textgreater \ 0=\ \textgreater \ x_1x_2-(x_1+x_2)+1\ \textgreater \ 0=\ \textgreater \ \\ \frac{m-1}{m+2} + \frac{2m}{m+2}+1\ \textgreater \ 0=\ \textgreater \ m\in(-\infty,-2)U(-1/4,+\infty) \\ In~final:m\in(-\infty,-2)U(-1/4,2) [/tex]