# How do you solve 9x^2 - 16 = 0?

Apr 20, 2016

$\left(3 x + 4\right) \left(3 x - 4\right)$

#### Explanation:

In the general form $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$

Treating $9 {x}^{2}$ as ${a}^{2}$
$\Rightarrow a = 3 x$
and
treating $16$ as ${b}^{2}$
$\Rightarrow b = 4$

$\textcolor{g r e e n}{\left(3 x + 4\right) \left(3 x - 4\right)} = 9 {x}^{2} - 16$