x² + x ≥ 0 x ( x +1 ) ≥ 0 daca x∈ ( - ∞ , -1] U [ 0, +∞ )
ridicam la patrat
x² + x + 2 = ( x² + x ) ² si notam t = x² +x
t + 2 = t²
t² - t - 2 = 0 Δ =1 +8 = 9 t₁ = -2/2 = -1 ; t₂ = 4 /2 = 2
atunci rez : x² + x = - 1 fals in R ; x² + x ≥ 0
x² + x = 2 ; x² + x - 2 = 0 ; Δ =9
x₁ = - 4 /2 = - 2 ; x₂ = 1
