Prosze o rozwiązanie
3x2 +15x + 24 >= 6
Po rozkladzie na postac iloczynowa badamy znaki i odp. Rysunek paraboli w zal.
a>0 ramiona paraboli ku gorze
3x²+15x+24 ≥ 6
3x²+15x+18 ≥ 0 /:3
x²+5x+6 ≥ 0
a = 1, b = 5, c = 6
Δ = b²-4ac = 25-24 = 1
√Δ = 1
x1 = (-b-√Δ)/2a = (-5-1)/2 = -3
x2 = (-b+√Δ)/2 = (-5+1)/2 = -2
a = 1 > 0, ramiona paraboli skierowane w górę
x ∈ (-∞,-3> u <-2,+∞)
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Po rozkladzie na postac iloczynowa badamy znaki i odp. Rysunek paraboli w zal.
a>0 ramiona paraboli ku gorze
3x²+15x+24 ≥ 6
3x²+15x+18 ≥ 0 /:3
x²+5x+6 ≥ 0
a = 1, b = 5, c = 6
Δ = b²-4ac = 25-24 = 1
√Δ = 1
x1 = (-b-√Δ)/2a = (-5-1)/2 = -3
x2 = (-b+√Δ)/2 = (-5+1)/2 = -2
a = 1 > 0, ramiona paraboli skierowane w górę
x ∈ (-∞,-3> u <-2,+∞)