Question

Решить уравнение (x^2+4x)^2+(x+2)^2 = 4

Answer 1

х^4+8х^3+16х^2+х^2+4х+4 = 4
х^4+8х^3+17х^2+4х = 0
х( х^3+8х^2+17х+4 ) = 0
х( х^3+4х^2+4х^2+16х+х+4 ) = 0
х( х^2(х+4)+4х(х+4)+1(х+4) ) = 0
х(х+4)(х^2+4х+1) = 0
х1 = 0
х2 = - 4
х^2+4х+1 = 0
D = 16 - 4 = 12
x3 = (-4+ 2 sqrt(3))/2 = - 2 + sqrt(3)
x4 = (-4- 2 sqrt(3))/2 = - 2 - sqrt(3)