Bonsoir,
a) 2 (x + 6)² - 9 = 0
2 (x² + 12x + 36) - 9 = 0
2x² + 24x + 63 = 0
a = 2 ; b = 24 ; c = 63
Δ = b² - 4ac
Δ = 24² - 4*2*63
Δ = 576 - 504
Δ = 72
Δ > 0 donc l'equation admet deux solutions
x1 = (-b - √Δ)/2a = (-24 - 6√2)/(2*2) = (-24 - 6√2)/4 = (-12 - 3√2)/2
x2 = (-b + √Δ)/2a = (-24 + 6√2)/(2*2) = (-24 + 6√2)/4 = (-12 + 3√2)/2
b) x² - 5 = 0
a = 1 ; b = 0 ; c = -5
Δ = b² - 4ac
Δ = 0² - 4*1*(-5)
Δ = 20
Δ > 0 donc l'equation admet deux solutions
x1 = (-b - √Δ)/2a = (-2√5)/2 = -√5
x2 = (-b + √Δ)/2a = (2√5)/2 = √5
c) (2x - 3)(5x + 1) = 0
Un produit de facteurs est nul si l'un de ses facteurs est nul
2x - 3 = 0
2x = 3
x = 3/2
5x + 1 = 0
5x = -1
x = -1/5
