Question

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

Answer 1

1)

${x}^{2} - 13x + 40 = 0$

$d = {13}^{2} - 4 \times 40 = 9$

$x1 = \frac{13 + 3}{2} = 8$

$x2 = \frac{13 - 3}{2} = 5$

2)

$6 {x}^{2} + x - 1 = 0$

$d = 1 + 4 \times 6 = 25$

$x1 = \frac{ -1 + 5}{12} = \frac{1}{3}$

$x2 = \frac{ - 1 - 5}{12} = - \frac{1}{2}$

3)

$x {}^{2} - 8x + 5 = 0$

$d = {8}^{2} - 4 \times 5 = 44$

$x1 = \frac{ 8+ 2\sqrt{11} }{2} = 4 + \sqrt{11}$

$x2 = \frac{8 - 2\sqrt{11} }{2} = 4 - \sqrt{11}$