Question

Найдите корни квадратного трехчлена:
a) x²+x-6;
6) 9²x-9x+2;
B) 0,2²x + 3x-20;
r) -2x²-х-0,125
1) 0,1x² +0,4;
e) -0,3x² +1,5х
помогите пожалуйста.
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Answer 1

a)x² + x - 6

a = 1

b = 1

c = -6

D = b² - 4ac = 1² - 4 × (-6) = 1 + 24 = 25

x1 = -b+кореньD/2a = -1+5/2 = 2

x2 = -b-кореньD/2a = -1-5/2 = -3

6)9x² - 9x + 2

a = 9

b = -9

c = 2

D = b² - 4ac = 9² - 4 × 2 × 9 = 81 - 79 = 9

x1 = -b+кореньD/2a = 9+3/18 = 2/3

x2 = -b-кореньD/2a = 9-3/18 = 1/3

B)0,2x² + 3x - 20

a = 0,2

b = 3

c = -20

D = b² - 4ac = 3² - 4 × 0,2 × (-20) = 9 + 16 = 25

x1 = -b+кореньD/2a = -3+5/0,4 = 5

x2 = -b-кореньD/2a = -3-5/0,4 = -20

r)-2x² - x - 0,125

a = -2

b = -1

c = -0,125

D = b² - 4ac = 1²- 4 × (-2) × (-0,125) = 1 - 1 = 0

x = -b/2a = 1/4

1)0,1x² + 0,4

a = 0,1

b = 0

c = 0,4

D = b² - 4ac = 0 - 4 × 0,1 × 0,4 = -0,16

e)-0,3x² + 1,5x

a = -0,3

b = 1,5

c = 0

D = b² - 4ac = 1,5² - 4 × (-0,3) × 0 = 1,5²

x1 = -b+кореньD/2a = -1,5+1,5/-0,6 = 0

x2 = -b-кореньD/2a = -1,5-1,5/-0,6 = 5