# How do you solve (2x-3)^2 = 8?

Sep 1, 2016

$x = 2.914 \mathmr{and} x = 0.0585$

#### Explanation:

Method 1

Find the square root of each side:

${\left(2 x - 3\right)}^{2} = 8$

$2 x - 3 = \pm \sqrt{8}$

Consider each root separately:

$2 x - 3 = \sqrt{8}$

$x = \frac{\sqrt{8} + 3}{2} = 2.914$

Or

$2 x - 3 = - \sqrt{8}$

$x = \frac{- \sqrt{8} + 3}{2} = 0.0858$

Method 2

Find the square of the binomial:

${\left(2 x - 3\right)}^{2} = 8$
$4 {x}^{2} - 12 x + 9 - 8 = 0 \leftarrow$ make equal to 0

$4 {x}^{2} - 12 x + 1 = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
x = (-(-12)+-sqrt((-12)^2-4(4)(1)))/(2(4)
$x = \frac{12 \pm \sqrt{128}}{8}$
$x = 2.914 \mathmr{and} x = 0.0858$