Решите уравнение (x+2) ⁴+2x²+8x-16=0

(x + 2) ^4 + 2 (x^2 + 4x - 8) = 0

(x + 2) ^4 + 2 (x^2 + 4x + 4 - 12) = 0

(x + 2) ^4 + 2 ((x + 2) ^2 - 12) = 0

(x + 2) ^4 + 2 (x + 2) ^2 - 24 = 0

Пусть (x + 2) ^2 = t, тогда

t^2 + 2t - 24 = 0

D = 4 + 4*24 = 100 = 10^2

t1 = ( - 2 + 10) / 2 = 8/2 = 4;

t2 = ( - 2 - 10) / 2 = - 12/2 = - 6;

1) (x + 2) ^2 = 4

(x + 2) ^2 - 2^2 = 0

(x + 2 - 2) (x + 2 + 2) = 0

x (x + 4) = 0

x1 = 0;

x2 = - 4

2) (x + 2) ^2 = - 6

нет решений

Ответ

- 4; 0