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

Feb 28, 2016

$x = - 1 \mathmr{and} x = - 3$

#### Explanation:

First multiply out to get in standard quadratic form :

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

$\therefore {x}^{2} + 4 x + 3 = 0$

Now factorize as a trinomial to obtain :

$\left(x + 1\right) \left(x + 3\right) = 0$

This implies that either $x + 1 = 0 \mathmr{and} x + 3 = 0$.

Hence $x = - 1 \mathmr{and} x = - 3$.