# How do you factor and solve 64x^2 - 1 = 0?

May 20, 2015

It can be done either using Bhaskara to find the roots or just go manipulating your equation, as it lacks the $b$ element - considering a quadratic as $a {x}^{2} + b x + c$.

In other words, let's do as follows to isolate $x$:

$64 {x}^{2} - 1 = 0$
$64 {x}^{2} = 1$
${x}^{2} = \frac{1}{64}$
$x = \sqrt{\frac{1}{64}}$
$x = \frac{\sqrt{1}}{\sqrt{64}}$
$x = \pm \frac{1}{8}$

In order to factor, you need to equal each root to zero.

You have your roots already: $\textcolor{g r e e n}{x = - \frac{1}{8}}$ and $\textcolor{red}{x = \frac{1}{8}}$.

Let's just equal each to zero:

$\textcolor{g r e e n}{8 x + 1 = 0}$
$\textcolor{red}{8 x - 1 = 0}$

Now, you know that

$64 {x}^{2} - 1 = \left(8 x + 1\right) \left(8 x - 1\right)$

May 20, 2015

$64 {x}^{2} = \left({8}^{2}\right) {x}^{2} = {\left(8 x\right)}^{2}$, so it is a perfect square.

$1 = {1}^{2}$, so it is also a perfect square.

$64 {x}^{2} - 1$ is a difference of squares, so it can be factored using:

${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

So,

$64 {x}^{2} - 1 = {\left(8 x\right)}^{2} - {\left(1\right)}^{2} = \left(8 x + 1\right) \left(8 x - 1\right)$

Now solving $\textcolor{w h i t e}{\text{sssssssss}}$ $64 {x}^{2} - 1 = 0$

is the same as solving $\left(8 x + 1\right) \left(8 x - 1\right) = 0$.

A product (multiply) of two numbers can be $0$ only if at least one of the numbers is $0$.

This tells us that to make $\left(8 x + 1\right) \left(8 x - 1\right) = 0$, we must make either:

$8 x + 1 = 0$ or $8 x - 0 = 0$

We can make $8 x + 1 = 0$, by $8 x = - 1$ so $x = - \frac{1}{8}$

And we can make $8 x - 1 = 0$ by $8 x = 1$ so $x = \frac{1}{8}$

The solutions are: $- \frac{1}{8}$ and $\frac{1}{8}$

This seems like a lot of work when you're just beginning, but with practice you'll write this:

$64 {x}^{2} - 1 = 0$

$\left(8 x + 1\right) \left(8 x - 1\right) = 0$

$8 x + 1 = 0$ $\textcolor{w h i t e}{\text{sss}}$ or $\textcolor{w h i t e}{\text{sss}}$ $8 x - 1 = 0$

$8 x = - 1$ $\textcolor{w h i t e}{\text{sss}}$ or$\textcolor{w h i t e}{\text{sss}}$ $8 x = 1$

$x = - \frac{1}{8}$ $\textcolor{w h i t e}{\text{sss}}$ or$\textcolor{w h i t e}{\text{sss}}$ $x = \frac{1}{8}$

The solutions are: $- \frac{1}{8}$ and $\frac{1}{8}$