# How do you solve (x+2)^2 = 9?

Jul 13, 2016

$x = 1$ or $x = - 5$

#### Explanation:

${\left(x + 2\right)}^{2} = 9$ can be written as ${\left(x + 2\right)}^{2} - 9 = 0$ or

${\left(x + 2\right)}^{2} - {3}^{2} = 0$

Now using the identity ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$, we have

$\left(x + 2 - 3\right) \left(x + 2 + 3\right) = 0$ or

$\left(x - 1\right) \left(x + 5\right) = 0$

Hence either $\left(x - 1\right) = 0$ i.e.

$x = 1$

or $\left(x + 5\right) = 0$ i.e. $x = - 5$