Question

Помогите пожалуйста ​

Answer 1

Ответ:

3.

${x}^{2} + 5x + 6 = 0 \\ d = {b}^{2} - 4ac = {25}^{2} - 4 \times 1 \times 6 = 25 - 24 = 1 \\ x1 = \frac{ - b - \sqrt{d} }{2a} = \frac{ - 5 - \sqrt{1} }{2 \times 1} = \frac{ - 5 - 1}{2} = \frac{ - 6}{2} = - 3 \\ x2 = \frac{ - b + \sqrt{d} }{2a} = \frac{ - 5 + \sqrt{1} }{2 \times 1} = \frac{ - 5 + 1}{2} = \frac{ - 4}{2} = - 2 \\ x1 + x2 = - 3 + ( - 2) = - 5 \\ x1 \times x2 = - 3 \times ( - 2) = 6$

4.

${x}^{2} + 13x - 3 = 0 \\ d = {b}^{2} - 4ac = {13}^{2} - 4 \times 1 \times ( - 3) = 169 + 12 = 181 \\ x1 = \frac{ - b - \sqrt{d} }{2a} = \frac{ - 13 - \sqrt{181} }{2} \\ x2 = \frac{ - b + \sqrt{d} }{2a} = \frac{ - 13 + \sqrt{181} }{2} \\ x1 + x2 + 4x1x2 = \frac{ - 13 - \sqrt{181} }{2} + \frac{ - 13 + \sqrt{181} }{2} + 4 \times \frac{ - 13 - \sqrt{181} }{2} \times \frac{ - 13 + \sqrt{181} }{2} = \\ \frac{ - 13 - \sqrt{181} - 13 + \sqrt{181} }{2} + 4 \times \frac{169 - 181}{4} = \frac{ - 26}{2} + ( - 12) = - 13 - 12 = - 25$

5.

$2 {x}^{2} + 10x = (x - 3)(x + 3) \\ 2 {x}^{2} + 10x = {x}^{2} - 9 \\ 2 {x}^{2} + 10x - {x}^{2} + 9 = 0 \\ {x}^{2} + 10x + 9 = 0 \\ d = {b}^{2} - 4ac = {10}^{2} - 4 \times 1 \times 9= 100 - 36 =64 \\ x1 = \frac{ - b - \sqrt{d} }{2a} = \frac{ - 10 - \sqrt{64} }{2} = \frac{ - 10 - 8}{2} = \frac{ - 18}{2} = - 9 \\ x2 = \frac{ - b + \sqrt{d} }{2a} = \frac{ - 10 + \sqrt{64} }{2} = \frac{ - 10 + 8}{2} = \frac{ - 2}{2} = - 1 \\ x1 + x2 = - 9 + ( - 1) = - 10 \\ x1 \times x2 = - 9 \times ( - 1) = 9$