Réponse :
Explications étape par étape
Exercice 1 :
a) x² + x - 6 = 0
⇔ x² + 3x - 2x - 6 = 0 factorisation par produit et somme
⇔ x ( x + 3 ) - 2 ( x + 3 ) = 0
⇔ ( x + 3 ) ( x - 2 ) = 0
x + 3 = 0 ou x - 2 = 0
⇔ x = - 3 ⇔ x = 2
S = { -3, 2 }
b) -2x² + x + 1 = 0
∆ = b²- 4 ac = 12 - 4· (-2) ·1 = 1 + 8 = 9
Δ>0
x₁ = -1 - √9 / 2 · (-2) = -1 - 3 / -4 = -4 / -4 = 1
x₂ = -1 + √9 / 2 · (-2) = -1 +3 / -4 = 2 / -4 = - 1/2
S = { -1/2 ; 1 }
c) x² + 4x + 5 = 0
∆ = b² - 4 ac = 42 - 4 ·1 · 5 = 16 - 20 = -4
Δ<0
Pas de solution
Exercice 2 :
h(x) = x² + 4x +5
⇔ h(x) = x² + 4x + 4 - 4 + 5
⇔ h(x) = ( x + 2 )² - 4 + 5
⇔ h(x) = ( x + 2 )² + 1
Exercice 3 :
g(x)= -2x²+ x + 1
g(x) = -2x² + x + 1 = 0
⇔ g(x) = -2x ( x - 1 ) - ( x - 1 ) = 0
⇔ g(x) = ( x - 1 ) ( -2x - 1 ) = 0
x-1 = 0 - 2x - 1 = 0
⇔ x = 1 ⇔ - 2x = 1
⇔2x = - 1
⇔ x = - 1/2
x -∞ -1/2 1 +∞
x - 1 - - 0 +
-2x - 1 + 0 - -
g(x) - 0 + 0 -
g(x) ≥ 0 sur [ 0 , 1 ]
g(x) ≤ 0 sur ] -∞ , -1/2 ] U [ 1 , + ∞ [
