# How do you solve 4x²-1=0?

Sep 3, 2016

1 or -1

#### Explanation:

$\left(2 x - 1\right) \left(2 x + 1\right) = 0$
$2 x - 1 = 0 , 2 x + 1 = 0$
$2 x = 0 + 1 , 2 x = 0 - 1$
$2 x = 1 , 2 x = - 1$
$x = \frac{1}{2} , x = - \frac{1}{2}$
$S . S \left\{+ \frac{1}{2} , - \frac{1}{2}\right\}$

Sep 3, 2016

$x = \pm \frac{1}{2}$

#### Explanation:

$\textcolor{g r e e n}{\text{You only need to solve as a quadratic if you have a } b x}$$\textcolor{g r e e n}{\text{term in the standard form of } y = a {x}^{2} + b x + c}$

Given:$\text{ } 4 {x}^{2} - 1 = 0$

$4 {x}^{2} = 1$
${x}^{2} = \frac{1}{4}$
$x = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2}$