# How do you solve the equation (4x-5)^2=64?

Feb 27, 2017

$x = \frac{13}{4}$

#### Explanation:

Square root both sides of the equation and simplify:
$\sqrt{{\left(4 x - 5\right)}^{2}} = \sqrt{64}$

$4 x - 5 = 8$

$4 x = 13$

$x = \frac{13}{4}$

Feb 27, 2017

$x = - \frac{3}{4} \text{ or } x = \frac{13}{4}$

#### Explanation:

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

$\sqrt{{\left(4 x - 5\right)}^{2}} = \textcolor{red}{\pm} \sqrt{64}$

$\Rightarrow 4 x - 5 = \textcolor{red}{\pm} 8$

•"solve "4x-5=color(red)(+)8

$4 x \cancel{- 5} \cancel{+ 5} = 8 + 5$

$\Rightarrow 4 x = 13$

divide both sides by 4

$\frac{\cancel{4} x}{\cancel{4}} = \frac{13}{4}$

$\Rightarrow x = \frac{13}{4}$

• "solve "4x-5=color(red)(-8)

$\Rightarrow 4 x - 5 + 5 = - 8 + 5$

$\Rightarrow 4 x = - 3$

$\Rightarrow x = - \frac{3}{4}$

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

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

${\left({\cancel{4}}^{1} \times \frac{13}{\cancel{4}} ^ 1 - 5\right)}^{2} = {8}^{2} = 64$

${\left({\cancel{4}}^{1} \times - \frac{3}{\cancel{4}} ^ 1 - 5\right)}^{2} = {\left(- 8\right)}^{2} = 64$

$\Rightarrow x = - \frac{3}{4} \text{ or "x=13/4" are the solutions}$