# How do you solve 3(x - 1)^2= 27?

Jun 21, 2018

$x = 10$

#### Explanation:

$3 {\left(x - 1\right)}^{2} = 27$

First, divide both sides by $\textcolor{b l u e}{3}$:
$\frac{3 {\left(x - 1\right)}^{2}}{\textcolor{b l u e}{3}} = \frac{27}{\textcolor{b l u e}{3}}$

${\left(x - 1\right)}^{2} = 9$

Now square root both sides:
$\sqrt{{\left(x - 1\right)}^{2}} = \sqrt{9}$

$x - 1 = \setminus \pm 3$

Finally, add $\textcolor{b l u e}{1}$ to both sides:
$x - 1 \quad \textcolor{b l u e}{+ \quad 1} = \setminus \pm 3 \quad \textcolor{b l u e}{+ \quad 1}$

Therefore,
$x = 4 \mathmr{and} - 2$

Hope this helps!

Jun 21, 2018

$x = - 2 \text{ or } x = 4$

#### Explanation:

$\text{divide both sides by 3}$

$\frac{\cancel{3}}{\cancel{3}} {\left(x - 1\right)}^{2} = 9$

${\left(x - 1\right)}^{2} = 9$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\sqrt{{\left(x - 1\right)}^{2}} = \pm \sqrt{9} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$x - 1 = \pm 3$

$\text{add 1 to both sides}$

$x = 1 \pm 3$

$x = 1 - 3 = - 2 \text{ or } x = 1 + 3 = 4$