Question #7c2d2

Jan 25, 2018

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

Explanation:

$9 {x}^{2} - 16 = 0$

Using the difference of squares, ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

$\therefore \left(3 x - 4\right) \left(3 x + 4\right) = 0$

We get two solutions:

$3 x - 4 = 0$, $3 x = 4$, $x = \frac{4}{3}$

$3 x + 4 = 0$, $3 x = - 4$, $x = - \frac{4}{3}$

Combining the two solutions, we can write

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

Jan 25, 2018

$9 {x}^{2} - 16 = 0$

Let's start by adding $\textcolor{red}{16}$ to both sides

$9 {x}^{2} \cancel{- 16} \cancel{\textcolor{red}{+ 16}} = 0 + \textcolor{red}{16}$

$9 {x}^{2} = 16$

Since we are solving for $x$, divide both sides by $9$

$\frac{\cancel{9} {x}^{2}}{\cancel{9}} = \frac{16}{9}$

${x}^{2} = \frac{16}{9}$

Now take square root

$x = \pm \sqrt{\frac{16}{9}}$

Simplify it

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