Question

Необходимо найти x. Должен быть дискриминант
(x-2)^4 + (x^2 - 4x) = 16

Answer 1

Ответ:

${( {(x - 2)}^{2} )}^{2} + ( {x}^{2} - 4x) = 16 \\ {(x {}^{2} - 4x + 4)}^{2} + ( {x}^{2} - 4x) = 16 \\ {x }^{2} - 4x = y \\ {(y + 4)}^{2} + y - 16 = 0 \\ {y}^{2} + 8y +16 + y - 16 = 0 \\ {y}^{2} + 9y = 0 \\ y(y + 9) = 0 \\ y = 0 \\ {x}^{2} - 4x = 0 \\ x(x - 4) = 0$

$x = 0 \\ x = 4 \\ y = - 9 \\ {x }^{2} - 4x = - 9 \\ {x}^{2} - 4x + 9 = 0 \\ d = 16 - 4 \times 9 = - 20$

нет решения

Ответ:х=0;х=4