# How do you solve (4x-1)^2= 16?

Oct 29, 2015

$x = - \frac{3}{4}$ and $x = \frac{5}{4}$

#### Explanation:

Take the square root of both sides

$\sqrt{{\left(4 x - 1\right)}^{2}} = \pm \sqrt{16}$

$4 x - 1 = \pm 4$

So we need to solve

$4 x - 1 = 4$ and $4 x - 1 = - 4$

$4 x = 5$

$x = \frac{5}{4}$

$4 x = - 3$

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