Odpowiedź:
zad 1
2x³- 9x² - 5x = 0
x(2x² - 9x -5) = 0
x = 0 ∨ 2x² - 9x - 5 = 0
2x² - 9x - 5 = 0
a = 2 , b = - 9 , c = - 5
Δ = b² - 4ac = (- 9)² - 4 * 2 * (- 5) = 81 + 40 = 121
√Δ = √121 = 11
x₁ = ( - b - √Δ)/2a = (9 - 11)/4 = - 2/4 = - 1/2
x₂ = ( - b + √Δ)/2a = (9 +11)/4 = 20/4 = 5
x = 0 ∨ x = - 1/2 ∨ x = 5
∨ - znaczy "lub"
zad 2
a)
(2x + 1)/(x - 1 ) = (x + 4)/(x - 3)
założenie:
x -1 ≠ 0 ∧ x - 3 ≠ 0
x ≠ 1 ∧ x ≠ 3
D: x ∈ R \ { 1 , 3 }
(2x + 1)/(x - 1 ) = (x + 4)/(x - 3)
(x - 3)(2x + 1) =(x - 1)(x + 4)
2x² -6x + x - 3 = x² - x + 4x - 4
2x² - 5x - 3 = x² + 3x - 4
2x² - x² - 5x - 3x - 3 + 4 = 0
x² - 8x + 1 = 0
a = 1 , b = - 8 , c = 1
Δ = b² - 4ac = (- 8)² - 4 * 1 * 1 = 64 - 4 = 60
√Δ = √60 = √(4 * 15) = 2√15
x₁ = ( - b - √Δ)/2a = (8 - 2√15)/2 = 2(4 - √15)/2 =4 -√15
x₂ = ( - b + √Δ)/2a = (8 +2√15)/2 =2(4 + √15)/2 =4 + √15
∧ - znaczy "i"
b)
(2x² +7x - 15)/(x² -25) = 0
założenie:
x² - 25 ≠ 0
(x - 5)(x + 5) ≠ 0
x - 5 ≠ 0 ∧ x + 5 ≠ 0
x ≠ 5 ∧ x ≠ - 5
x ∈ R \ { - 5 , 5 }
2x² + 7x - 15 = 0
a = 2 , b = 7 , c = - 15
Δ = b² - 4ac = 7² - 4 * 2 * (- 15) = 49 + 120 = 169
√Δ = √169 = 13
x₁ = ( - b - √Δ)/2a = (- 7 - 13)/4 = - 20/4 = - 5 nie należy do dziedziny
x = ( - b + √Δ)/2a = (- 7 + 13)/4 = 6/4 = 1 2/4 = 1 1/2
Odp: x = 1 1/2 = 1,5
