How do you solve ${(x + 2)}^{2} = 1$ $(x + 2)^2 = 1$ ?

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Answer 1 · 2016-02-28T19:37:13

$x = - 1 or x = - 3$ $x=-1 or x=-3$

First multiply out to get in standard quadratic form :

$x^{2} + 4 x + 4 = 1$ $x^2+4x+4=1$

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

Now factorize as a trinomial to obtain :

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

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

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

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