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

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 =========================== 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 ===========================

- 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