# How do you solve the equation 5(x-4)^2=125?

May 5, 2017

$x = - 1 \text{ or } x = 9$

#### Explanation:

$\textcolor{b l u e}{\text{Isolate " (x-4)^2" by dividing both sides by 5}}$

$\frac{\cancel{5} {\left(x - 4\right)}^{2}}{\cancel{5}} = \frac{125}{5}$

$\Rightarrow {\left(x - 4\right)}^{2} = 25$

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

$\sqrt{{\left(x - 4\right)}^{2}} = \textcolor{red}{\pm} \sqrt{25} \leftarrow \text{ note plus or minus}$

$\Rightarrow x - 4 = \pm 5$

$\text{add 4 to both sides}$

$x \cancel{- 4} \cancel{- 4} = \pm 5 + 4$

$\Rightarrow x = 4 \pm 5$

$\Rightarrow x = 4 + 5 = 9 \text{ or } x = 4 - 5 = - 1$

$\textcolor{b l u e}{\text{As a check}}$

Substitute these values into the left side of the equation and if equal to the right side then they are the solutions.

$x = - 1 \to 5 {\left(- 1 - 4\right)}^{2} = 5 \times {5}^{2} = 125$

$x = 9 \to 5 {\left(9 - 4\right)}^{2} = 5 \times {5}^{2} = 125$

$\Rightarrow x = - 1 \text{ or " x=9" are the solutions}$