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