Răspuns:
Explicație pas cu pas:
ax²+2(3a-1)x+a+3 = 0
ecuatia are solutii reale distincte <=> Δ > 0
Δ = (6a-2)²-4a·(a+3)
Δ = 36a²-24a+4-4a²-12a
Δ = 32a²-36a+4 = 4·(8a²-9a+1) =>
8a²-9a+1 > 0
8a²-9a+1 = 0 => a₁,₂ = [9±√(81-32)]/16 = (9±7)/16
a₁ = 1/8 ; a₂ = 1
a I -∞ 1/8 1 +∞
8a²-9a+1 I++++++++0--------0++++++++ =>
a ∈ (-∞ , 1/8) ∪ (1 , +∞)
