Question

Прошу помощь с алгеброй

240. Решите уравнение:

a) (x - 2)2 = 2(3x-10);

б) (3x + 2)2 = (3x + 2);

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Answer 1

a)

${x}^{2} - 4x + 4 = 6x - 20 \\ {x}^{2} - 4x + 4 - 6x + 20 = 0 \\ {x}^{2} - 10x + 24 = 0 \\ d = 100 - 4 \times 24 \times 1 = 100 - 96 = \\ = 4 = {2}^{2}$

$x1 = \frac{10 + 2}{2} = \frac{12}{2} = 6 \\ x2 = \frac{10 - 2}{2} = \frac{8}{2} = 4$

б)

$9 {x}^{2} + 12x + 4 - 3x - 2 = 0 \\ {9x}^{2} + 9x + 2 = 0 \\ d = 81 - 4 \times 9 \times 2 = 81 - 72 = \\ = 9 = {3}^{2} \\ x1 = \frac{ - 9 + 3}{18} = - \frac{6}{18} = - \frac{1}{3}$

$x2 = \frac{ - 9 - 3}{18} = - \frac{12}{18} = - \frac{2}{3}$