Question

Rozwiąż 3 równania

a) x³-5x-4=0

b)x³-3x+2=0

c)x⁴-7x²+6x=0

Wskazówka: w podpunkcie a) zapisz -5x jako -x-4x

Pozdrawiam i z góry dzięki za rozwiązania :)

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Answer 1

a)

x³ - 5x - 4 = 0

x³ - x - 4x - 4 = 0

x·(x² - 1) - 4·(x + 1) = 0

x·(x - 1)(x + 1) - 4·(x + 1) = 0

(x + 1)[x·(x - 1) - 4] = 0

(x + 1)(x² - x - 4) = 0

x + 1 = 0  ∨  x² - x - 4 = 0

 

x + 1 = 0

x = - 1

 

x² - x - 4 = 0

 

Odp.

 

b)

x³ - 3x + 2 = 0

x³ - x - 2x + 2 = 0

x·(x² - 1) - 2·(x - 1) = 0

x·(x - 1)(x + 1) - 2·(x - 1) = 0

(x - 1)[x·(x + 1) - 2] = 0

(x - 1)(x² + x - 2) = 0

x - 1 = 0  ∨ x² + x - 2 = 0

 

x - 1 = 0

x = 1

 

x² + x - 2 = 0

Δ = 1² - 4 · 1 · (- 2) = 1 + 8 = 9; √Δ = 3

x₁ = ⁻¹ ⁻ ³/₂·₁ = ⁻⁴/₂ = - 2

x₂ = ⁻¹ ⁺ ³/₂·₁ = ²/₂ = 1

 

Odp. x = - 2 lub x = 1

 

c)

x⁴ - 7x² + 6x = 0

x·(x³ - 7x + 6) = 0

x·(x³ - x - 6x + 6) = 0

x·[x·(x² - 1) - 6·(x - 1)] = 0

x·[x·(x - 1)(x + 1) - 6·(x - 1)] = 0

x·{(x - 1)[x·(x + 1) - 6]} = 0

x·(x - 1)(x² + x - 6) = 0

x = 0  ∨  x - 1 = 0  ∨  x² + x - 6 = 0

 

x = 0

 

x - 1 = 0

x = 1

 

x² + x - 6 = 0

Δ = 1² - 4 · 1 · (- 6) = 1 + 24 = 25; √Δ = 5

x₁ = ⁻¹⁻⁵/₂·₁ = ⁻⁶/₂ = - 3

x₂ = ⁻¹⁺⁵/₂·₁ = ⁴/₂ = 2

 

Odp. x = - 3 lub x = 0 lub x = 1 lub x = 2