Odpowiedź:
zad 1
f(x) = x²
x² = 0
x = 0
Jedno miejsce zerowe x = 0
zad 2
f(x) = x² - 1
x² - 1 = 0
(x - 1)(x + 1) = 0
x - 1 = 0 ∨ x + 1 = 0
x = 1 ∨ x = - 1
zad 3
f(x) = x² - 9
x² - 9 = 0
(x - 3)(x + 3) = 0
x - 3 = 0 ∨ x + 3 = 0
x = 3 ∨ x = - 3
zad 4
f(x) = x² - 2
x² - 2 = 0
(x - √2)(x + √2) = 0
x - √2 = 0 ∨ x + √2 = 0
x = √2 ∨ x = - √2
zad 5
f(x) = 81x² - 25
81x² - 25 = 0
(9x - 5)(9x + 5) = 0
9x - 5 = 0 ∨ 9x + 5 = 0
9x = 5 ∨ 9x = - 5
x = 5/9 ∨ x = - 5/9
zad 6
- 4x² + 16 = 0
- 4(x² - 4) = 0
x² - 4 = 0
(x - 2)(x + 2) = 0
x - 2 = 0 ∨ x + 2 = 0
x = 2 ∨ x = - 2
zad 7
f(x) = x² - 4/9
x² - 4/9 = 0
(x - 2/3)(x + 2/3) = 0
x - 2/3 = 0 ∨ x + 2/3 = 0
x = 2/3 ∨ x = - 2/3
zad 8
f(x) = x² - 3/25
x² - 3/25 = 0
(x - √3/5)(x + √3/5) = 0
x - √3/5 = 0 ∨ x + √3/5 = 0
x = √3/5 ∨ x = - √3/5
zad 9
f(x) = x² + 1
Ponieważ x² + 1 > 0 dla x ∈ R , więc funkcja nie ma miejsc zerowych
zad 10
f(x) = 9x² + 1
Ponieważ 9x² + 1 > 0 dla x ∈ R , więc funkcja nie ma miejsc zerowych
zad 11
f(x) = - 4x² + 1
- 4x² + 1 = 0
a = - 4 , b = 0 , c = 1
Δ = b² - 4ac = 0² - 4 * (- 4) * 1 = 16
√Δ = √16 = 4
x₁ = (- b - √Δ)/2a = (0 - 4)/(- 8) = - 4/(- 8) = 4/8 = 1/2
x₂ = (- b + √Δ)/2a = (0 + 4)/(- 8) = - 4/8 = - 1/2
