# How do you solve 9x^2-4=0?

Oct 21, 2017

$x = \setminus \pm \setminus \frac{2}{3}$

#### Explanation:

$9 {x}^{2} - 4 = 0$

Move the $4$ to the RHS

$9 {x}^{2} = 4$

Divide both sides by $9$

${x}^{2} = \setminus \frac{4}{9}$

Take the square root of both sides

$\setminus \sqrt{{x}^{2}} = \setminus \pm \setminus \sqrt{\setminus \frac{4}{9}}$

Simplify

$x = \setminus \pm \setminus \frac{2}{3}$

$\setminus \therefore x = - \setminus \frac{2}{3} , \setminus q \quad \setminus q \quad \setminus q \quad x = \setminus \frac{2}{3}$

Oct 21, 2017

x= +-sqrt(4/9

In decimal $x = \pm 0.67$

#### Explanation:

$9 {x}^{2} - 4 = 0$
Let start by adding $\textcolor{red}{4}$ to both sides
$9 {x}^{2} \cancel{- 4} \cancel{\textcolor{red}{+ 4}} = 0 + \textcolor{red}{4}$
$9 {x}^{2} = 4$
Divide both sides by $\textcolor{g r e e n}{9}$
$\frac{\cancel{9} {x}^{2}}{\cancel{\textcolor{g r e e n}{9}}} = \frac{4}{\textcolor{g r e e n}{9}}$
${x}^{2} = \frac{4}{9}$
We need to take square root sqrt
$x = \pm \sqrt{\frac{4}{9}}$