[tex]ax^2+2(a+1)x+2a-1 \geq0 \\ \\ $Inecuatia NU are solutie in multimea reala. \\ Astfel, noi trebuie sa punem conditiile ca inecuatia sa aibe solutii \\ in semnul opus, adica, cand este $ \ \textless \ 0. \\ \\ ax^2+2(a+1)x+2a-1 \ \textless \ 0 \\ \\$Consideram: $f(x) = ax^2+2(a+1)x+2a-1 \\ \\ $Stim ca daca: \\ \\ \boxed{1 }\quad \left\{ \begin{array}{c} a'\ \textless \ 0 \\ \Delta \ \textless \ 0 \end{array} \right |\Rightarrow f(x)\ \textless \ 0 \\ \\[/tex]
[tex] \boxed{2} \quad \left\{ \begin{array}{c} a'\ \textgreater \ 0 \\ \Delta \ \textless \ 0 \end{array} \right |\Rightarrow f(x) \ \textgreater \ 0 \\ \\ \\ $Noi conditia \boxed{1} trebuie sa o punem: \\ \\ \left\{ \begin{array}{c} a'\ \textless \ 0 \\ \Delta \ \textless \ 0 \end{array} \right \Rightarrow \left\{ \begin{array}{c} a \ \textless \ 0 \\ b^2-4a'c \ \textless \ 0 \end{array} \right \Rightarrow
\left\{ \begin{array}{c} a \ \textless \ 0 \\ 4(a+1)^2-4a(2a-1)\ \textless \ 0 \end{array} \right \Rightarrow
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[tex]\Rightarrow \left\{ \begin{array}{c} a \ \textless \ 0 \\ 4(a^2+2a+1)-4a(2a-1)\ \textless \ 0 \end{array} \right \Rightarrow \\ \\ \Rightarrow \left\{ \begin{array}{c} a \ \textless \ 0 \\ 4\Big(a^2+2a+1-a(2a-1)\Big)\ \textless \ 0 \end{array} \right \Rightarrow \\ \\ \Rightarrow \left\{ \begin{array}{c} a \ \textless \ 0 \\ 4(a^2+2a+1-2a^2+a)\ \textless \ 0 \end{array} \right \Rightarrow \left\{ \begin{array}{c} a \ \textless \ 0 \\ 4(-a^2+3a+1)\ \textless \ 0\Big|:(-4) \end{array} \right \Rightarrow [/tex]
[tex]\Rightarrow \left\{ \begin{array}{c} a \ \textless \ 0 \\ a^2-3a-1\ \textgreater \ 0 \end{array} \right \Rightarrow \left\{ \begin{array}{c} a \in (-\infty,0) \\ a\in \Big(-\infty, \dfrac{3-\sqrt{13}}{2}\Big) \cup \Big(\dfrac{3+\sqrt{13}}{2},\infty\Big) \end{array} \right | \\ \\ \\ $Din cele doua conditii din sistem intersectate rezulta ca:$ \\\\ \Rightarrow \boxed{\boxed{a\in\Big(-\infty, \dfrac{3-\sqrt{13}}{2}\Big)}} $ - (solutie finala)[/tex]
