How do you solve #(x + 2)^2 = 1 #?

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Answer 1 · 2016-02-28T19:37:13

#x=-1 or x=-3#

First multiply out to get in standard quadratic form :

#x^2+4x+4=1#

#therefore x^2+4x+3=0#

Now factorize as a trinomial to obtain :

#(x+1)(x+3)=0#

This implies that either #x+1=0orx+3=0#.

Hence #x=-1 or x=-3#.