Найти корни квадратного уравнения x2 -5x+6 = 06x2 - 7x – 3 = 04х2 – 13х + 9 = 02х2 + 7х...

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Найти корни квадратного уравнения x2 -5x+6 = 06x2 - 7x – 3 = 04х2 – 13х + 9 = 02х2 + 7х + 5 = 05х2 + 3х -2 = 0​


Алгебра (12 баллов) | 32 просмотров
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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

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