Odpowiedź:
b)
(x + 7)² > 0
x²+ 14x + 49 > 0
Obliczamy miejsca zerowe
x² + 14x+ 49 = 0
a = 1 , b =14 , c = 49
Δ = b² - 4ac = 14² - 4 * 1 * 49 = 196 - 196 = 0
x₁ = x₂ = - b/2a = - 14/2 = - 7
x ∈ R \ { - 7}
e)
(x + 2)(x - 3) < 6
x² + 2x - 3x - 6 - 6 < 0
x² - x - 12 < 0
Obliczamy miejsca zerowe
x² - x- 12 = 0
a = 1 , b = - 1 , c = - 12
Δ = b² - 4ac = (- 1)² - 4 * 1 * (- 12) = 1 + 48 = 49
√Δ = √49 = 7
x₁ = ( - b - √Δ)/2a = (1 - 7)/2 =- 6/2 = - 3
x₂ = ( - b + √Δ)/2a = (1 + 7)/2 = 8/2 = 4
(x + 3)(x - 4) < 0
x + 3 > 0 ∧ x - 4 < 0 ∨ x + 3 < 0 ∧ x - 4 > 0
x > - 3 ∧ x < 4 ∨ x < - 3 ∧ x > 4
x > - 3 ∧ x < 4
x ∈ ( - 3 , 4 )
