x² - (2a-6)x + 3a + 9 = 0
x1 < 0, x2 > 0
Ne folosim de relatiile lui Viete:
Conditia 1) x1+x2 ≤ 0 sau x1+x2 ≥ 0 => x1+x2 ∈ R => a ∈ R
Conditia 2) x1·x2 < 0 => c/a < 0 => 3a+9 < 0 => 3a<-9 => a < -3
Conditia 3) Δ > 0 => (2a-6)² - 4(3a+9) > 0 => 4a²-24a+36 - 12a -36 > 0 =>
=> 4a²-36 > 0 => a² - 9 > 0 => a² > 9 => a ∈ (-∞,-3) ∪ (3, +∞)
Din (1) ∩ (2) ∩ (3) => a ∈ (-∞,-3)
