Question

sa se determine a∈ Rpentru care inecuatia ax²+2(a+1)x+2a-1>=o nu are solutie in multimea reala

Answer 1

ax² + 2(a+1)x+ 2a - 1 ≥ 0  nu are soluție în ℝ ⇒

⇒ ax² + 2(a+1)x+ 2a - 1 < 0 ,  ∀x∈ ℝ⇒ a< 0  și Δ < 0

Δ < 0  ⇒ (a + 1)² -a(2a - 1) < 0 ⇒ a² - 3a - 1 > 0⇒
 
⇒a∈ (-∞,  (3-√13)/2) ∪ ((3+√13)/2,  ∞) 

Dar, a < 0 ⇒ a∈ (-∞,  (3-√13)/2)

Answer 2

$ax^2+2(a+1)x+2a-1 \geq0 \\ \\ Inecuatia NU are solutie in multimea reala. \\ Astfel, noi trebuie sa punem conditiile ca inecuatia sa aibe solutii \\ in semnul opus, adica, cand este \ \textless \ 0. \\ \\ ax^2+2(a+1)x+2a-1 \ \textless \ 0 \\ \\ Consideram: f(x) = ax^2+2(a+1)x+2a-1 \\ \\ Stim ca daca: \\ \\ \boxed{1 }\quad \left\{ \begin{array}{c} a'\ \textless \ 0 \\ \Delta \ \textless \ 0 \end{array} \right |\Rightarrow f(x)\ \textless \ 0$

[tex] \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 [/tex]

$\Rightarrow \left\{ \begin{array}{c} a \ \textless \ 0 \\ 4(a^2+2a+1)-4a(2a-1)\ \textless \ 0 \end{array} \right \Rightarrow \\ \\ \Rightarrow \left\{ \begin{array}{c} a \ \textless \ 0 \\ 4\Big(a^2+2a+1-a(2a-1)\Big)\ \textless \ 0 \end{array} \right \Rightarrow \\ \\ \Rightarrow \left\{ \begin{array}{c} a \ \textless \ 0 \\ 4(a^2+2a+1-2a^2+a)\ \textless \ 0 \end{array} \right \Rightarrow \left\{ \begin{array}{c} a \ \textless \ 0 \\ 4(-a^2+3a+1)\ \textless \ 0\Big|:(-4) \end{array} \right \Rightarrow$

$\Rightarrow \left\{ \begin{array}{c} a \ \textless \ 0 \\ a^2-3a-1\ \textgreater \ 0 \end{array} \right \Rightarrow \left\{ \begin{array}{c} a \in (-\infty,0) \\ a\in \Big(-\infty, \dfrac{3-\sqrt{13}}{2}\Big) \cup \Big(\dfrac{3+\sqrt{13}}{2},\infty\Big) \end{array} \right | \\ \\ \\ Din cele doua conditii din sistem intersectate rezulta ca: \\\\ \Rightarrow \boxed{\boxed{a\in\Big(-\infty, \dfrac{3-\sqrt{13}}{2}\Big)}} - (solutie finala)$