rozwiąż nierówność 3x²+2x>1
3x2 + 2x > 1
3x2 + 2x +1 > 0
delta = b2 - 4ac
delta = 4-4*3 = 4-12 = -8
delta ujemna
miejsce zerowe = -b/2a = -2/6 = -1/3
x= -1/3
3x²+2x-1>0
Δ=b²-4ac=4+12=16
x₁=[-b-√Δ]/2a=[-2-4]/6=-1
x₂=[-b+√Δ]/2a=[-2+4]/6=⅓
x∈(-∞;-1) ∨ (⅓;+∞)
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3x2 + 2x > 1
3x2 + 2x +1 > 0
delta = b2 - 4ac
delta = 4-4*3 = 4-12 = -8
delta ujemna
miejsce zerowe = -b/2a = -2/6 = -1/3
x= -1/3
3x²+2x-1>0
Δ=b²-4ac=4+12=16
x₁=[-b-√Δ]/2a=[-2-4]/6=-1
x₂=[-b+√Δ]/2a=[-2+4]/6=⅓
x∈(-∞;-1) ∨ (⅓;+∞)