# How do you solve 16x^2 - 81 = 0 by factoring?

Apr 30, 2018

$x = - \frac{9}{4} , \frac{9}{4}$

#### Explanation:

Use the rule for difference of squares.
$16 {x}^{2} - 81 = 0$
$\left(4 x - 9\right) \left(4 x + 9\right) = 0$

This equation will be true if either (4x-9) or (4x+9) is 0.
$4 x + 9 = 0$
$4 x = - 9$
$x = - \frac{9}{4}$
Or
$4 x - 9 = 0$
$4 x = 9$
$x = \frac{9}{4}$

$x = - \frac{9}{4} , \frac{9}{4}$

Aug 10, 2018

$x = \pm \frac{9}{4}$

#### Explanation:

Recall that this is a difference of squares which factors as

bar ul(|color(white)(2/2)a^2-b^2=(a+b)(a-b)color(white)(2/2)|

Both of our terms are perfect squares, where our $a = 4 x$ and $b = 9$. This allows us to factor this as

$\left(4 x + 9\right) \left(4 x - 9\right) = 0$

We can set both factors equal to zero to get

$4 x + 9 = 0 \implies 4 x = - 9 \implies x = - \frac{9}{4}$ and

$4 x - 9 = 0 \implies 4 x = 9 \implies x = \frac{9}{4}$

Hope this helps!