Question

$12x^{2}+bx-119=0\\x_{1} =-2\frac{1}{3}$ Уравнение. Найти 2 корень.

Answer 1

$12 {x}^{2} + bx - 119 = 0 \\ D = {b}^{2} + 12 \times 4 \times 119 = \\ = {b}^{2} + 5712 \\ \\ x_1 = \frac{ - b + \sqrt{ {b}^{2} + 5712} }{24} = - 2 \frac{1}{3} \\ x_2 = \frac{ - b - \sqrt{ {b}^{2} + 5712 } }{24}$

решим уравнение

$\frac{ \sqrt{ {b}^{2} + 5712} - b}{24} = - \frac{7}{3} \: \: \: | \times 24 \\ \sqrt{ {b}^{2} + 5712 } - b = - 56 \\ \sqrt{ {b}^{2} + 5712 } = b - 56 \: \: \: \\ {b}^{2} + 5712 = {(b - 56)}^{2} \\ {b}^{2} + 5712 = {b}^{2} - 112 b + 3136 \\ 112b = 3136 - 5712 \\ 112b = - 2576 \\ b = - 23$

Найдем второй корень

$x_1 = \frac{ - b - \sqrt{ {b}^{2} + 5712} }{24} = \frac{ 23 - \sqrt{529 + 5712} }{24} = \\ = \frac{ 23 - 79}{24} = - \frac{56}{24} = - \frac{7}{3} \\ x_2 = \frac{ - b + \sqrt{ {b}^{2} + 5712} }{24} = \frac{23 + 79}{24} = \frac{102}{24} = 4.25$

ответ: 4,25