Rozwiąż nierównośc -2x²+7x-3>0
-2x² + 7x - 3 > 0
Δ = b²-4ac
x₁ = -b-√Δ/2a
x₂ = -b+√Δ/2a
Δ = 7² - 4 x (-2) x (-3) = 25
√Δ = √25 = 5
x₁ = -7-5/2*(-2) = 3
x₂ = -7+5/2*(-2) = 0,5
Miejsca zerowe funkcji to 0,5 i 3; ramiona paraboli w dół
-2x² + 7x - 3 > 0 <=> x ∈ (0,5; 3)
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Pozdrawiam :)
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-2x² + 7x - 3 > 0
Δ = b²-4ac
x₁ = -b-√Δ/2a
x₂ = -b+√Δ/2a
Δ = 7² - 4 x (-2) x (-3) = 25
√Δ = √25 = 5
x₁ = -7-5/2*(-2) = 3
x₂ = -7+5/2*(-2) = 0,5
Miejsca zerowe funkcji to 0,5 i 3; ramiona paraboli w dół
-2x² + 7x - 3 > 0 <=> x ∈ (0,5; 3)
----------------------------------------------------------------------------------------------------------
Feci, quod potui, faciant meliora potentes
Pozdrawiam :)