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Foxyck
@Foxyck
August 2022
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Помогите решить! Очень срочно!
Разложите на множители квадратный трехчлен.
1) 5a^2+19a-4
2) 7c^2+20c-3
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Answers & Comments
krugly1981
1) 5a² + 19a - 4 = 0
D = b² - 4ac = 19² - 4 × 5 × (-4) = 361 + 80 = 441 = 21²
x1 = ( - 19 + 21) / 10 = 0,2
x2 = ( - 19 - 21) / 10 = - 4
Разложим по формуле:
a( x - x1)(x - x2)
5(x - 0,2)(x + 4)
2) 7c² + 20c - 3 = 0
D = b² - 4ac = 20² - 4 × 7 ×(-3) = 400 + 84 = 484 = 22²
x1 = ( - 20 + 22) / 14 = 2/14 = 1/7
x2 = ( - 20 - 22) / 14 = - 42/14 = - 3
7( с - 1/7)( c + 3)
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Answers & Comments
D = b² - 4ac = 19² - 4 × 5 × (-4) = 361 + 80 = 441 = 21²
x1 = ( - 19 + 21) / 10 = 0,2
x2 = ( - 19 - 21) / 10 = - 4
Разложим по формуле:
a( x - x1)(x - x2)
5(x - 0,2)(x + 4)
2) 7c² + 20c - 3 = 0
D = b² - 4ac = 20² - 4 × 7 ×(-3) = 400 + 84 = 484 = 22²
x1 = ( - 20 + 22) / 14 = 2/14 = 1/7
x2 = ( - 20 - 22) / 14 = - 42/14 = - 3
7( с - 1/7)( c + 3)