Równania kwadratowe
Podaj dwie kolejne liczby naturalne, których iloczyn jest równy 756.
Liczby:
n
n+1
n∈N
n(n+1)=756
n²+n-756=0
Δ=1+3024=3025
√Δ=55
n₁=(-1+55)/2 = 27
n₂=(-1-55)/2= -28 -> nie jest naturalna
Odp.: Szukane liczby to 27 oraz 28.
x * (x + 1) = 756x² + x = 756x² + x - 756 = 0--------------------a = 1 b = 1 c = -756Δ = b² * 4acΔ = 1² - 4 * 1* (-756)Δ = 1 + 3024 = Δ = 3025√Δ = √(3025)Δ = 55x₁ = -b - √Δ/2a x₁ = (-1 - 55)/2x₁ = -56/2 = x₁ = -28 x₂ = -b + √Δ x₂ = (-1 + 55)/2 x₂ = 54/2 x₂ = 27Liczby te to 27 i 28========================Masz pytania to pisz na pw..
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Liczby:
n
n+1
n∈N
n(n+1)=756
n²+n-756=0
Δ=1+3024=3025
√Δ=55
n₁=(-1+55)/2 = 27
n₂=(-1-55)/2= -28 -> nie jest naturalna
Odp.: Szukane liczby to 27 oraz 28.
x * (x + 1) = 756
x² + x = 756
x² + x - 756 = 0
--------------------
a = 1 b = 1 c = -756
Δ = b² * 4ac
Δ = 1² - 4 * 1* (-756)
Δ = 1 + 3024 =
Δ = 3025
√Δ = √(3025)
Δ = 55
x₁ = -b - √Δ/2a
x₁ = (-1 - 55)/2
x₁ = -56/2 =
x₁ = -28
x₂ = -b + √Δ
x₂ = (-1 + 55)/2
x₂ = 54/2
x₂ = 27
Liczby te to 27 i 28
========================
Masz pytania to pisz na pw..