Odpowiedź:
A)
f(x) = 2x² - 5x + 3
2x² - 5x + 3 = 0
a = 2 , b = - 5 , c = 3
Δ = b² - 4ac = (- 5)² - 4 * 2 * 3 = 25 - 24 = 1
√Δ = √1 = 1
x₁ = (- b - √Δ)/2a = ( 5 - 1)/4 = 4/4 = 1
x₂ = (- b + √Δ)/2a = (5 + 1)/4 = 6/4 = 1 2/4 = 1 1/2
B)
f(x) = - x² +2x - 1
- x² + 2x -1 = 0
a = - 1 , b = 2 , c = - 1
Δ = b² - 4ac = 2² - 4 * (- 1) * (- 1) = 4 - 4 = 0
x₁ = x₂ = - b/2a = - 2/(- 2) = 2/2 = 1
C)
f(x) = 3x² - x + 3
3x² - x + 3 = 0
a = 3 , b = - 1 , c = 3
Δ = b² - 4ac = (- 1)² - 4 * 3 * 3 = 1 - 36 = - 35
Δ < 0 , więc brak miejsc zerowych
