Question

Найдите корни квадратного трехчлена:
1)x^2+3x-10
2) x^2-13x+40
3)6x^2+x-1
приравняйте к нулю и решите.

Answer 1

Ответ:

1.(x-2)(x+5)

2.(x-8)(x-5)

3.(3x-1)(2x+1)

Пошаговое объяснение:

1.(x-2)(x+5)

2.(x-8)(x-5)

3.(3x-1)(2x+1)

Answer 2

${x}^{2} + 3x - 10 = 0 \\ d = 9 - 4 \times 1 \times ( - 10) = 9 + 40 = 49 = {7}^{2} \\ x1 = \frac{ - 3 + 7}{2} = 2 \\ x2 = \frac{ - 3 - 7}{2} = - 5 \\ {x}^{2} - 13x + 40 = 0 \\ d = 169 - 4 \times 1 \times 40 = 169 - 160 = 9 = {3}^{2} \\ x1 = \frac{13 + 3}{2} = 8 \\ x2 = \frac{13 - 3}{2} = 5 \\ 6 {x}^{2} + x - 1 = 0 \\ d = 1 - 4 \times 6 \times ( - 1) = 1 + 24 = 25 = {5}^{2} \\ x1 = \frac{ - 1 + 5}{12} = \frac{4}{12} = \frac{1}{3} \\ x2 = \frac{ - 1 - 5}{12} = \frac{ - 6}{12} = \frac{ - 1}{2} = - 0.5$