Réponses

  • Omnes
  • Modérateur confirmé
2014-05-09T19:42:59+02:00
Salut,

2 x²+ x - 6 ≥ x + 2
2x² + x - 6 -x - 2
≥0
2x² - 8
≥ 0
2(x² - 4) ≥0
2(x-2)(x+2) ≥0
x-2 ≥0
x ≥ 2

x+2≥0
x≥-2

S = [-2;+infini[

et
2 x²+ x - 6 = -6

2x² + x - 6 + 6 = 0
2x² + x = 0
x(2x + 1) = 0
x = 0
2x + 1 = 0
2x = -1
x = -1/2

S = {-1/2;0}.

Bonne soirée !