Odpowiedź:
a)
3x - y + 1 = 0 → y = 3x + 1
y = x² + 2x + 3
3x + 1 = x² +2x + 3
x² + 2x +3 - 3x - 1 = 0
x² - x + 2 = 0
a = 1 , b = - 1 , c = 2
Δ = b² - 4ac = (- 1)² - 4 * 1 * 2 = 1 - 8 = - 7
Δ < 0 więc brak miejsc zerowych
x ∈ ∅ (zbiór pusty)
c)
- 4y = x² + 4x + 8
1/8x - 1/4y = - 1 | * 8
4y = - x² - 4x - 8 |: 4
x - 2y = - 8
y = - 1/4x² - x - 2
- 2y = - x - 8
y = - 1/4x² - x - 2
2y = x + 8 → y = 1/2x + 4
- 1/4x² - x - 2 = 1/2x + 4
- 1/4x² - x - 1/2x - 2 - 4 = 0
- 1/4x² - (3/2)x - 6 = 0 | * 4
- x² - (3 * 2)x - 6 = 0
- x² - 6x - 6 = 0
a = - 1 , b = - 6 , c = - 6
Δ = b² - 4ac = (- 6)² - 4 * (- 1) * (- 6) = 36 - 24 = 12
√Δ = √12 = √(4 * 3) = 2√3
x₁ = (- b - √Δ)/2a = ( 6 - 2√3)/(- 2) = - 2(√3 - 3)/(- 2) = √3 - 3
x₂ = (- b + √Δ)/2a = (6 + 2√3)/(- 2) = 2(3 + √3)/(- 2) = - ( 3 + √3)
