Odpowiedź:
Szczegółowe wyjaśnienie:
3.
a) x²- x -6 <0
Δ = 1 + 24 = 25
√Δ = 5
x1 =(1+5)/2 = 3
x2 =(1-5)/2 = -2
x∈ (-2 ; 3)
b) x²+6x+8 ≥ 0
Δ= 36-32 = 4
√Δ = 2
x1 = (-6+2)/2 = -2
x2 = (-6-2)/2 = -4
x∈ (-∞ ; -4> lub <-2 ; +∞)
c) 2x²+3x-2 ≤ 0
Δ = 9 + 16 = 25
√Δ = 5
x1 = (-3+5)/4 = 1/2
x2 = (-3-5)/4 = -2
x∈ <-2 ; 1/2>
d) 3x²+5x+2 > 0
Δ = 25 - 24 = 1
√Δ = 1
x1 = (-5+1)/6 = -2/3
x2 = (-5-1)/6 = -1
x∈ (-∞ ; -1) lub (-2/3 ; +∞)
e) -x²+3x-2 ≥ 0 /*(-1)
x²-3x+2 ≤ 0
Δ = 9 - 8 =1
√Δ = 1
x1 = (3+1)/2 = 2
x2 = (3-1)/2 = 1
x∈ <1 ; 2>
f) -2x²+10x-6 < 0 /*(-1)
2x²- 10x +6 > 0
Δ = 100 - 48 = 52
√Δ = √52 = 2√13
x1 = (10+2√13)/4 = (5+√13)/2
x2 = (10-2√13)/4 = (5-√13)2
x∈ (-∞ ; (5-√13)/2) lub ( (5+√13)/2 ; +∞)
