Odpowiedź:
1)
x⁴ - 5x² + 6 = 0
za x² wstawiam z
z² - 5z + 6 = 0
a = 1 , b = - 5 , c = 6
Δ = b² - 4ac = (- 5)² - 4 * 1 * 6 = 25 - 24 = 1
√Δ = √1 = 1
z₁ = ( - b - √Δ)/2a = (5 - 1)/2 = 4/2 = 2
z₂ = ( - b + √Δ)/2a = (5 + 1)/2 = 6/2 = 3
x₁² = 2
x₂² = 3
(x₁ - √2)(x₁ + √2) = 0
x₁ - √2 = 0 ∨ x₁ + √2 = 0
x₁ = √2 ∨ x₁ = - √2
(x₂² - √3)(x₂² + √3) = 0
x₂ = √3 ∨ x₂ = - √3
Mamy 4 rozwiązania
x = √2 ∨ x = - √2 ∨ x = √3 ∨ x = - √3
2)
9x⁴ = 12x² - 4
9x⁴ - 12x² + 4 = 0
za x wstawiam z
9x² - 12z + 4 = 0
a = 9 , b = - 12 , c = 4
Δ = b² - 4ac = (- 12)² - 4 * 9 * 4 = 144 - 144 = 0
z₁ = z₂ = - b/2a = 12/18 = 2/3
x₁² = 2/3
[x₁ - √(2/3)][(x₁ + √(2/3)] = 0
x₁ = √(2/3) ∨ x₁ = - √(2/3)
√(2/3) = √2/√3 = (√2 * √3)/3 = √6/3
x₁ = √6/3 ∨ x₁ = - √6/3
Mamy 2 rozwiązania
x = √6/3 ∨ x = - √6/3
