Question

ПОМОГИТЕ
Решите уравнение
(x+3)^4+5(x+3)^2-36=0

Answer 1

$(x+3)^4+5(x+3)^2-36=0 \ (x+3)^2=t \ t^2+5t-36=0 \ D=b^2-4ac\ D=25-4*1*36=25+36*4=169 \ t_{1,2} = frac{-5б13}{2} \ t_1 = 4 ; t_2 = - 9 \ (x+3)^2 neq -9 \ (x+3)^2 = 4 \ (x+3)(x+3)=4 \ x^2+3x+3x+9-4=0 \ x^2+6x+5=0 \ D=36-5*4*1=16 \ x_{1,2} = frac{-6б4}{2} \ x_1 = -1 ; x_2 = -5$