Question

Помогите

Знайдіть корені квадратного тричлена:
1)x^2-13x+40
2)6x^2+x-1
3)x^2-8x+5

Answer 1

Ответ:

1)8, 5

2)1/3, -1/2

3)4+√11, 4-√11

Объяснение:

$1)d = {( - 13)}^{2} - 4 \times 1 \times 40 = \\ = 169 - 160 = 9 \\ x1 = \frac{13 + \sqrt{9} }{2 \times 1} = \frac{13 + 3}{2} = \\ = \frac{16}{2} = 8 \\ x2 = \frac{13 - \sqrt{9} }{2 \times 1} = \frac{13 - 3}{2} = \\ = \frac{10}{2} = 5$

$3)d = {( - 8)}^{2} - 4 \times 1 \times 5 = \\ = 64 - 20 = 44 \\ x1 = \frac{ 8+ \sqrt{44} }{2} = \frac{ 8 + 2 \sqrt{11} }{2} = \\ = 4 + \sqrt{11} \\ x2 = \frac{ 8 - \sqrt{44} }{2 } = \frac{8 - 2\sqrt{11} }{2} = \\ = 4 - \sqrt{11}$

$2)d = {1}^{2} - 4 \times 6 \times ( - 1) = \\ = 1 + 24 = 25 \\ x1 = \frac{ - 1 + \sqrt{25} }{2 \times 6} = \frac{ - 1 + 5}{12} \\ = \frac{4}{12} = \frac{1}{3} \\ x2 = \frac{ - 1 - \sqrt{25} }{2 \times 6} = \frac{ - 1 - 5}{12} \\ = \frac{ - 6}{12} = - \frac{1}{2}$