Question

найти корни квадратного уравнения

x2 -5x+6 = 0
6x2 - 7x – 3 = 0
4х2 – 13х + 9 = 0
2х2 + 7х + 5 = 0
5х2 + 3х -2 = 0
​

Answer 1

Ответ:

${x}^{2} - 5x + 6 = 0 \\ d = 25 - 24 = 1 \\ x1 = \frac{5 + 1}{2} = \frac{6}{2} = 3$

$x2 = \frac{5 - 1}{2} = \frac{4}{2} = 2$

$6 {x}^{2} - 7x - 3 = 0 \\ d = 49 + 4 \times 6 \times 3 = 49 + 72 = \\ = 121 = {11}^{2}$

$x1 = \frac{7 + 11}{12} = \frac{18}{12} = 1 \frac{6}{12} = 1 \frac{1}{2} \\ x2 = \frac{7 - 11}{12} = - \frac{4}{12} = - \frac{1}{3}$

$4 {x}^{2} - 13x + 9 = 0 \\ d = 169 - 4 \times 4 \times 9 = 169 - 144 = \\ = 25 = {5}^{2}$

$x1 = \frac{13 + 5}{10} = \frac{18}{10} = 1 \frac{8}{10} = 1.8 \\ x2 = \frac{13 - 5}{10} = \frac{8}{10} = 0.8$

${2x}^{2} + 7x + 5 = 0 \\ d = 49 - 40 = 9 = {3}^{2} \\ x1 = \frac{ - 7 - 3}{4} = - \frac{10}{4} = - 2 \frac{2}{4} = \\ = - 2.5$

$x2 = \frac{ - 7 + 3}{4} = - \frac{4}{4} = - 1$

$5 {x}^{2} + 3x - 2 = 0 \\ d = 9 + 40 = 49 = {7}^{2} \\ x1 = \frac{ - 3 + 7}{10} = \frac{4}{10} = 0.4$

$x2 = \frac{ - 3 - 7}{10} = \frac{ - 10}{10} = - 1$