Question

5. Найдите нули квадратичной функции:
1) у = х2 - x;
3) y = 12х - 17х +6;
5) y = 3х - 5х + 8; срочно пжпжпжпж​

Answer 1

1)⠀y = x² - x
x² - x = 0
x(x - 1) = 0
x₁ = 0 ⠀ x - 1 = 0
⠀⠀⠀⠀⠀ x₂ = 1

3)⠀y = 12x² - 17x + 6
12x² - 17x + 6 = 0

D = 289 - 288 = 1
x = (17 ± 1) / 24

x₁ = 16 / 24 = 2/3
x₂ = 18 / 24 = 3/4


5)⠀y = 3x² - 5x + 8
3x² - 5x + 8 = 0

D = 25 - 96 = - 71

D < 0⠀=>⠀x ∉ ℝ