Odpowiedź:
zad 1
- x² + (m + 2)x + 8m - 1 < 0
a = - 1 , b = m + 2 , c = 8m - 1
założenie:
a < 0 ∧ Δ < 0
Δ = b² - 4ac = (m + 2)² - 4 * (- 1) * (8m - 1) = m² + 4m + 4 + 32m - 4 =
= m² + 36m
m² + 36m < 0
m(m + 36) < 0
m < 0 ∧ m + 36 > 0 ∨ m > 0 ∧ m + 36 < 0
m < 0 ∧ m > - 36
m ∈ ( - 36 , 0 )
zad 2
f(x) = √[kx² - (k + 1)x - 2k + 3]
kx² - (k + 1)x - 2k + 3 ≥ 0
a = k , b = - (k + 1) , c = - 2k + 3
założenie :
a > 0 ∧ Δ ≤ 0
k > 0
Δ = b² - 4ac = [- (k + 1)]² - 4 * k * (- 2k + 3) = k² + 2k + 1 + 8k² - 12k =
= 9k² - 10k + 1
9k² - 10k + 1 ≤ 0
Δ = (- 10)² - 4 * 9 * 1 = 100 - 36 = 64
√Δ = √64 = 8
k₁ = ( - b - √Δ)/2a = (10 - 8)/18 = 2/18 = 1/9
k₂ = ( - b + √Δ)/2a = (10 + 8)/18 = 18/18 = 1
