a) f(x) = -4
(x + 1)² + x - 4 = -4
(x + 1)² + x = 0
x² + 2x + 1 + x = 0
x² + 3x + 1 = 0
On calcule Δ = b² - 4ac = 3² - 4×1×1 = 9
Δ > 0 donc 2 solutions
x1 = (-b - √Δ)/(2a) = (-3 - √5) / 2
x2 = (-b + √Δ)/(2a) = (-3 + √5) / 2
b) f(x) = x
(x + 1)² + x - 4 = x
(x + 1)² - 4 = 0
[(x+1) - 2][(x+1) + 2] = 0 (identité remarquable)
(x - 1)(x + 3) = 0
Produit nul :
Soit x - 1 = 0 et x = 1
Soit x + 3 = 0 et x = -3
Solutions : 1 et -3
c) f(x) = x - 4
(x + 1)² + x - 4 = x - 4
(x + 1)² = 0
x + 1= 0
x = -1
Solution : -1
Voilà ! :)
