Question

If x^2 + y^2 + 4x + 4y + 8 = 0, then (x + y) is equal to​

Answer 1

1. Answer:
2. here is your answer

According to the question:

According to the question:x^2+y^2+4x+4y+8=0

According to the question:x^2+y^2+4x+4y+8=0=) x^2+y^2+4x+4y+4+4=0

According to the question:x^2+y^2+4x+4y+8=0=) x^2+y^2+4x+4y+4+4=0=)(x^2+4x+4)+(y^2+4y+4)=0

According to the question:x^2+y^2+4x+4y+8=0=) x^2+y^2+4x+4y+4+4=0=)(x^2+4x+4)+(y^2+4y+4)=0=){(x^2)+4x+(2^2)}+{(y^2+4y+(2^2)}=0

According to the question:x^2+y^2+4x+4y+8=0=) x^2+y^2+4x+4y+4+4=0=)(x^2+4x+4)+(y^2+4y+4)=0=){(x^2)+4x+(2^2)}+{(y^2+4y+(2^2)}=0=)(x+2)^2+(y+2)^2=0 [because a^2+2ab+b^2 =(a+b)^2]

According to the question:x^2+y^2+4x+4y+8=0=) x^2+y^2+4x+4y+4+4=0=)(x^2+4x+4)+(y^2+4y+4)=0=){(x^2)+4x+(2^2)}+{(y^2+4y+(2^2)}=0=)(x+2)^2+(y+2)^2=0 [because a^2+2ab+b^2 =(a+b)^2]=)(x+2)^2=0 and (y+2)^2=0 [because square of any number is greater than or equal to zero]

According to the question:x^2+y^2+4x+4y+8=0=) x^2+y^2+4x+4y+4+4=0=)(x^2+4x+4)+(y^2+4y+4)=0=){(x^2)+4x+(2^2)}+{(y^2+4y+(2^2)}=0=)(x+2)^2+(y+2)^2=0 [because a^2+2ab+b^2 =(a+b)^2]=)(x+2)^2=0 and (y+2)^2=0 [because square of any number is greater than or equal to zero]=)x+2=0 and y+2=0

According to the question:x^2+y^2+4x+4y+8=0=) x^2+y^2+4x+4y+4+4=0=)(x^2+4x+4)+(y^2+4y+4)=0=){(x^2)+4x+(2^2)}+{(y^2+4y+(2^2)}=0=)(x+2)^2+(y+2)^2=0 [because a^2+2ab+b^2 =(a+b)^2]=)(x+2)^2=0 and (y+2)^2=0 [because square of any number is greater than or equal to zero]=)x+2=0 and y+2=0=)x= -2 and y= -2

According to the question:x^2+y^2+4x+4y+8=0=) x^2+y^2+4x+4y+4+4=0=)(x^2+4x+4)+(y^2+4y+4)=0=){(x^2)+4x+(2^2)}+{(y^2+4y+(2^2)}=0=)(x+2)^2+(y+2)^2=0 [because a^2+2ab+b^2 =(a+b)^2]=)(x+2)^2=0 and (y+2)^2=0 [because square of any number is greater than or equal to zero]=)x+2=0 and y+2=0=)x= -2 and y= -2Therefore, x+y = (-2)+ (-2)

According to the question:x^2+y^2+4x+4y+8=0=) x^2+y^2+4x+4y+4+4=0=)(x^2+4x+4)+(y^2+4y+4)=0=){(x^2)+4x+(2^2)}+{(y^2+4y+(2^2)}=0=)(x+2)^2+(y+2)^2=0 [because a^2+2ab+b^2 =(a+b)^2]=)(x+2)^2=0 and (y+2)^2=0 [because square of any number is greater than or equal to zero]=)x+2=0 and y+2=0=)x= -2 and y= -2Therefore, x+y = (-2)+ (-2)=-2–2

According to the question:x^2+y^2+4x+4y+8=0=) x^2+y^2+4x+4y+4+4=0=)(x^2+4x+4)+(y^2+4y+4)=0=){(x^2)+4x+(2^2)}+{(y^2+4y+(2^2)}=0=)(x+2)^2+(y+2)^2=0 [because a^2+2ab+b^2 =(a+b)^2]=)(x+2)^2=0 and (y+2)^2=0 [because square of any number is greater than or equal to zero]=)x+2=0 and y+2=0=)x= -2 and y= -2Therefore, x+y = (-2)+ (-2)=-2–2=-4

According to the question:x^2+y^2+4x+4y+8=0=) x^2+y^2+4x+4y+4+4=0=)(x^2+4x+4)+(y^2+4y+4)=0=){(x^2)+4x+(2^2)}+{(y^2+4y+(2^2)}=0=)(x+2)^2+(y+2)^2=0 [because a^2+2ab+b^2 =(a+b)^2]=)(x+2)^2=0 and (y+2)^2=0 [because square of any number is greater than or equal to zero]=)x+2=0 and y+2=0=)x= -2 and y= -2Therefore, x+y = (-2)+ (-2)=-2–2=-4Therefore, x+y= -4

Answer 2

Answer:

2x or 2y

Step-by-step explanation:

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

(x+2)^2=(y+2)^2

x+2=y+2

x=y

x+y=x+x or y+y=2x or 2y