# How do you solve (y - 8) ^ { 2} = 49?

Jun 25, 2017

$y = 1 \text{ or } y = 15$

#### Explanation:

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

$\sqrt{{\left(y - 8\right)}^{2}} = \pm \sqrt{49} \leftarrow \textcolor{red}{\text{ note plus or minus}}$

$\Rightarrow y - 8 = \pm 7$

$\text{add 8 to both sides}$

$y \cancel{- 8} \cancel{+ 8} = 8 \pm 7$

$\Rightarrow y = 8 \pm 7$

$y = 8 + 7 \Rightarrow y = 15 \leftarrow \text{ is a solution}$

$y = 8 - 7 \Rightarrow y = 1 \leftarrow \text{ is a solution}$

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

$y = 15 \to {\left(15 - 8\right)}^{2} = {7}^{2} = 49 \leftarrow \text{ true}$

$y = 1 \to {\left(1 - 8\right)}^{2} = {\left(- 7\right)}^{2} = 49 \leftarrow \text{ true}$

Jun 25, 2017

15 or 1

#### Explanation:

Each side is a perfect square, so we just need to square root each side, so ${\left(y - 8\right)}^{2} = 49$ becomes $y - 8 = \text{+/-} 7$. All numbers need to be put onto one side so $y - 8 = \text{+/-} 7$ becomes $y = 8 \text{+/-} 7$.

$y = 8 + 7 = 15$
or
$y = - 7 + 8 = 1$.