Question

Solve. (x + 2)2 = 1

x = –3 and x = –1

x = 3 and x = 1

x = –1 and x = 4

x = –4 and x = –3

Answer 1

\[(x+2)^2=1\]\[x^2+4x+4=1\]\[x^2+4x+4-1=0\]\[x^2+4x+3=0\]

Answer 2

ok now what

Answer 3

\[x^2+x+3x+3=0\]\[x(x+1)+3(x+1)=0\]\[(x+3)(x+1)=0\]
x=-3 or x=-1

Answer 4

understood??

Answer 5

@harrisonshields simply get the square root of both sides as such:
\[\sqrt{(x+2)^{2}} = \sqrt{1}\]
\[x+2 = \pm 1\]
so from here
x+2 =1+or
x+2 = -1
can you proceed from here?