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2009-12-04T16:36:38+01:00
-4k²-16x+9=0

a = -4
b = -16
c = 9
Δ = b² -4ac
Δ = (-16)² -4*(-4) *9 = 256 + 144 = 400
√Δ = √400 = 20
k₁ = (-b-√Δ): (2*a)
k₁ = [-(-16) - 20] :2*(-4) = (16 -20): (-8) = (-4) :(-8) = 0,5
k₂= (-b+√Δ): (2*a)
k₂ = [-(-16) +20] :2*(-4) = (16 +20): (-8) = ( 36) :(-8) = - 4,5

2(2x-3)(x+1)-5(x-1)²= 2(x-2)(x-1)
2(2x² +2x -3x -3) -5(x² -2x +1) = 2(x² -x -2x + 2)
4x² -2x -6 -5x² +10x -5 = 2x² -6x +4
-x² + 8x - 11 = 2x² -6x +4
-x² -2x² +8x +6x -11-4 = 0
-3x² +14x -15 = 0
a = -3
b = 14
c = -15
Δ = b² -4ac
Δ = 14² -4*(-3)*(-15) = 196 -180 = 16
√Δ = √16 = 4
x₁ = (-b-√Δ): (2*a)
x₁ = (-14-4):[2*(-3)] = (-18): (-6) = 3
x₂= (-b+√Δ): (2*a)
x₂ = (-14+4):[2*(-3)] = (-10): (-6)= 5/3

-x²-3x+4≥0
a= -1
b = -3
c = 4
Δ = 9 +16 =25
√Δ = 5

x₁ = (-2):(-2) = 1
x₂ = 8:(-2) = -4
Parabola skierowana est ramionami w dół
x ∈ < -5, 1>


x²-7x+12>0
a = 1
b = -7
c = 12
Δ = (-7)² -4*1*12= 49 -48 = 1
√Δ = 1
x₁ = 6 : 2 = 3
x₂ = 8 : 2 = 4
Ramiona paraboli skierowane są do góry
x ∈ (-∞, 3) ∨( 4 +∞)

-4a²-16a+9<0
a = -4
b = -16
c = 9
Δ = (-16)² -4*(-4)*(-16) = 256 -256 = 0
√Δ = 0
a₁ = 16 : (-8) = -2
a₂ = 16 : (-8) = -2 podwójny pierwiastek

Ramiona paraboli skierowane są w dół
a ∈ R - { -2}


x²-6x+9≤0

a = 1
b = -6
c = 9
Δ = (-6)² -4*1*9= 36 -36 = 0
√Δ = 0
x₁ = x₂ = (-b): 2a
x₁ = x₂ = 6 : 2*1 = 6 : 2 = 3
Ramiona paraboli skierowane są w górę
x = 3