Odpowiedź:
w(x) = (x - 2)(x² + 5x + 6)(x² + 4) = 0
Ponieważ x² + 4 > 0 dla x ∈ R , więc :
(x - 2)(x² + 5x + 6) = 0
x - 2 = 0 ∨ x² + 5x + 6 = 0
x = 2
x² + 5x + 6 = 0
a = 1 , b = 5 , c = 6
Δ = b² - 4ac = 5² - 4 * 1 * 6 = 25 - 24 = 1
√Δ = √1 = 1
x₁ = ( - b - √Δ)/2a = ( - 5 - 1)/2 = - 6/2 = - 3
x₂ = (- b + √Δ)/2a = (- 5 + 1)/2 = - 4/2 = - 2
x = { - 3 , - 2 , 2 }
Odp: C
