# How do you solve (x - 4)^2 = 16?

Mar 29, 2018

$x = 0 , 8$

#### Explanation:

Given: ${\left(x - 4\right)}^{2} = 16$

Square root both sides:

$x - 4 = \sqrt{16}$

$x - 4 = \pm 4$

So we got: ${x}_{1} - 4 = 4 , {x}_{2} - 4 = - 4$.

${x}_{1} = 4 + 4$

$= 8$

${x}_{2} = - 4 + 4$

$= 0$

$\therefore x \setminus \in \left\{0 , 8\right\}$