Question

(4+х)в кубе -х в квадрате(х-12)=16

Answer 1

$(4+x)^3 -x^2*(x-12)=16 (4+x)^3-x^3+12x^2-16=0 64+48x+12x^2+x^3-x^3+12x^2-16=0 24x^2+48x+48=0 D=2304-4608=-2304 x_1= frac{-48+i sqrt{2304} }{2*24} = frac{-48+i*48}{48} =-1+i x_2= frac{-48-i sqrt{2304} }{2*24} = frac{-48-i*48}{48} =-1-i$