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m1dway
@m1dway
July 2022
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Найдите корни квадратного трехчлена:
1. x^2+12x+27
2.8x^2-13-6
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Universalka
Verified answer
1) x² + 12x + 27 = 0
D = 12² - 4*27 = 144 - 108 = 36
X1,2 = (- 12 + - √36)/2 = (- 12 + - 6)/2
X1 = (- 12 + 6)/2 = - 3
X2 = (- 12 - 6)/2 = - 9
Ответ : - 3; - 9
2) 8x² - 13x - 6 = 0
D = 13² - 4*(-6)*8 = 169 + 192 =361
X1,2 = (13 + - √361)/16 = (13 + - 19)/16
X1 = (13 + 19)/16 = 32/16 = 2
X2 = (13 - 19)/16 = - 6/16 = - 0, 375
Ответ: - 0,375; 2
2 votes
Thanks 1
KristinSerdg
1)D=12 (в квадрате) -4*27=корень из 36
x(1)=-12+6/2=-3
x(2)=-12-6/2=-9
2)D=(-13в квадрате)-4*8*(-6)=корень из 361
x(1)=13+19/16=0,5
x(2)=13-19/16=-3/8
0 votes
Thanks 0
Universalka
Во втором задании X1 = 2, а не 0,5
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Answers & Comments
Verified answer
1) x² + 12x + 27 = 0D = 12² - 4*27 = 144 - 108 = 36
X1,2 = (- 12 + - √36)/2 = (- 12 + - 6)/2
X1 = (- 12 + 6)/2 = - 3
X2 = (- 12 - 6)/2 = - 9
Ответ : - 3; - 9
2) 8x² - 13x - 6 = 0
D = 13² - 4*(-6)*8 = 169 + 192 =361
X1,2 = (13 + - √361)/16 = (13 + - 19)/16
X1 = (13 + 19)/16 = 32/16 = 2
X2 = (13 - 19)/16 = - 6/16 = - 0, 375
Ответ: - 0,375; 2
x(1)=-12+6/2=-3
x(2)=-12-6/2=-9
2)D=(-13в квадрате)-4*8*(-6)=корень из 361
x(1)=13+19/16=0,5
x(2)=13-19/16=-3/8