Question

Помогите, пожалуйста с этим номером. ​

Answer 1

Ответ:

Решай через дискриминант, что там сложного?

Answer 2

Ответ:

$1) {x}^{2} + 5x + 4 = 0 \\ d = {b}^{2} - 4ac = {5}^{2} - 4 \times 1 \times 4 = 25 - 16 = 9 \\ x1 = \frac{ - b - \sqrt{d} }{2a} = \frac{ - 5 - \sqrt{9} }{2 \times 1} = \frac{ - 5 - 3}{2} = \frac{ - 8}{2} = - 4 \\ x2 = \frac{ - b + \sqrt{d} }{2a} = \frac{ - 5 + \sqrt{9} }{2 \times 1} = \frac{ - 5 + 3}{2} = \frac{ - 2}{2} = - 1 \\ {x}^{2} + 5x + 4 = (x + 4)(x + 1)$

$2) {x}^{2} - 5x + 4 = 0 \\ d = {b}^{2} - 4ac = {( - 5)}^{2} - 4 \times 1 \times 4 = 25 - 16 = 9 \\ x1 = \frac{ - b - \sqrt{d} }{2a} = \frac{ - ( - 5) - \sqrt{9} }{2 \times 1} = \frac{5 - 3}{2} = \frac{2}{2} = 1 \\ x2 = \frac{ - b + \sqrt{d} }{2a} = \frac{ - ( - 5) + \sqrt{9} }{2 \times 1} = \frac{ 5 + 3}{2} = \frac{ 8}{2} = 4 \\ {x}^{2} - 5x + 4 = (x - 1)(x - 4)$

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

$4) {a}^{2} + 4ab + 3 {b}^{2} = {a}^{2} + ab + 3ab + 3 {b}^{2} = a(a + b) + 3b(a + b) = (a + 3b)(a + b)$