Question

(4y^2+8y+8)^(1/2)+(3y^2+6y+12)^(1/2)=4-2y-y^2
Помогите решить уравнение

Answer 1

 $sqrt{4y^2+8y+8}+ sqrt{3y^2+6y+12}=4-2y-y^2\\ sqrt{4(y^2+2y+2)}+sqrt{3(y^2+2y+4)} = -(y^2+2y-4)\\ y^2+2y=t\\ sqrt{4(t+2)}+sqrt{3(t+4)} = 4-t\\ 7t+20+2sqrt{12(t+2)(t+4)}=(4-t)^2\\ 2sqrt{12(t+2)(t+4)}=t^2-15t-4\\ 4*12(t+2)(t+4)=(t^2-15t-4)^2\\ (23-t)(t-4)^2(t+1)=0\\ t=23\ t=4\ t=-1$     
 
 $y^2+2y=23\ y=-2sqrt{6}-1\ y=2sqrt{6}-1\\ y^2+2y=4\ y=-sqrt{5}-1\ y=sqrt{5}-1\\ y^2+2y=-1\ y=-1$ 
Подходит    $y=-1$