zadanie w załaczniku daje naj
1) 5x^2 - 6x + 1 = 0Δ = 36 - 20 = 16√Δ = 4x1 = 6 + 4 / 10 = 1
x2 = 6 -4 / 10 = 1/5f(x) = 5(x-1)(x-1/5)2) x^2 + 6x + 9Δ = 36 - 36 = 0x1 = 6/2 = 3f(x) = (x - 3)^23) x^2 - x +5 = 0 Δ = 1 - 20 mniejsza od zera, nie ma postaci iloczynowej4) 2x^2 - 8 = 0x^2 = 4x1 = 2x2 = -2f(x) = 2(x-2)(x+2)
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1) 5x^2 - 6x + 1 = 0
Δ = 36 - 20 = 16
√Δ = 4
x1 = 6 + 4 / 10 = 1
x2 = 6 -4 / 10 = 1/5
f(x) = 5(x-1)(x-1/5)
2) x^2 + 6x + 9
Δ = 36 - 36 = 0
x1 = 6/2 = 3
f(x) = (x - 3)^2
3) x^2 - x +5 = 0
Δ = 1 - 20 mniejsza od zera, nie ma postaci iloczynowej
4) 2x^2 - 8 = 0
x^2 = 4
x1 = 2
x2 = -2
f(x) = 2(x-2)(x+2)