Question

Rozwiąż równania
a) 2x²-10x-12=0
b) 3x³+3x²-12x-12=0

Answer 1

A)
2 x² - 10 x - 12 = 0 / : 2
x² - 5 x - 6 = 0
Δ = 25 -4*1*(-6) = 25 + 24 = 49
√Δ = 7
x1 = [5 - 7]/2 = -1
x2 = [5 + 7]/2 = 6
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b)
3x³ + 3x² - 12 x - 12 = 0
3x²*(x +1) - 12*(x +1) = 0
(3 x² -12)*(x+1) = 0
3*(x² - 4)*(x+1) = 0
3*(x -2)*(x +2)*(x +1) = 0
x1 = -2, x2 = -1, x3 = 2
===========================

Answer 2

- to jest równanie kwadratowe więc wylicza się: Δ, x₁ i x₂

a) 2x²-10x-12=0

a = 2
b = -10
c = -12

Δ = b² - 4ac
Δ = 100 - 4 * 2 * -12
Δ = 100 + 96 (minus razy minus daje+)
Δ = 196
√Δ = 14

x₁ = -b - √Δ / 2a = 10 - 14 / 4 = -4/4 = -1
x₂ = -b + √Δ / 2a = 10 + 14 / 4 = 24/4 = 6


b)

3x³ + 3x² - 12 x - 12 = 0
3x² * ( x + 1 ) - 12 * ( x + 1 ) = 0
( 3x² - 12 ) * ( x + 1 ) = 0
3 * ( x² - 4 ) * ( x + 1 ) = 0
3 *( x - 2 ) * ( x + 2 ) * ( x + 1 ) = 0
x₁ = -2
x₂ = -1
x₃ = 2