Odpowiedź:
a)
x⁶ + x =x + 2x⁴
x⁶- 2x⁴ + x - x = 0
x⁶ - 2x⁴ = 0
x⁴(x² - 2) = 0
x⁴ = 0 ∨ x² - 2 = 0
x = 0 ∨ x = √2 ∨ x = - √2
∨ - znaczy "lub"
b)
1/6x³ + 2x² + 6x = 0 | * 6
x³ + 12x² + 36x = 0
x(x² + 12x + 36) = 0
x = 0 ∨ x² + 12x + 36 = 0
x² + 12x + 36 =0
a = 1 , b = 12 , c = 36
Δ = b² - 4ac = 12² - 4 * 1 * 36 = 144 - 144 = 0
x₁ = x₂ = - b/2a = - 12/2 = - 6
x = 0 ∨ x = - 6
c)
2x² - 3x = x - x³
x³ + 2x² - 3x - x = 0
x³ + 2x² - 4x = 0
x(x² - 2x - 4) = 0
x = 0 ∨ x² - 2x - 4 = 0
x² - 2x - 4 = 0
a = 1 , b = - 2 , c = - 4
Δ = b² - 4ac = (- 2)² - 4 * 1 * (- 4) = 4 + 16 = 20
√Δ = √20 = √(4 * 5) = 2√5
x₁ = ( - b - √Δ)/2a = ( 2 - 2√5)/2 = 2(1 - √5)/2 = 1 - √5
x₂ = (- b + √Δ)/2a = (2 + 2√5)/2 = 2(1 + √5)/2 = 1 + √5
x = 0 ∨ x = 1 - √5 ∨ x = 1 + √5
d)
x³ + 4x = - 5x²
x³ + 5x² + 4x = 0
x(x² + 5x + 4) = 0
x = 0 ∨ x² + 5x + 4 = 0
x² + 5x + 4 = 0
a = 1 , b = 5 , c = 4
Δ = b² - 4ac = 5² - 4 * 1 * 4 = 25 - 16 = 9
√Δ = √9 = 3
x₁ = (- b - √Δ)/2a = ( -5 - 3)/2 = - 8/2 = - 4
x₂ =(- b + √Δ)/2a = (- 5 + 3)/2 = - 2/2 = - 1
x = 0 ∨ x = - 4 ∨ x = - 1
