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  • Roma
  • Community Manager
2010-01-23T19:26:17+01:00
A)
x² - 4x < 0
Δ = 16 - 0 = 16
√Δ = 4
x₁ = 4 - 4 / 2 = 0
x₂ = 4 + 4 / 2 = 4
x ∈ (0, 4)

b)
x² - 3x + 2 ≥ 0
Δ = 9 - 8 = 1
x₁ = 3 - 1 / 2 = 1
x₂ = 3 + 1 / 2 = 2
x ∈ (-∞, 1] u [2, +∞)

c)-x² + 4x - 4 ≥ 0
Δ = 16 - 16 = 0
x = -4 / -2 = 2
x ∈ {2}