Question

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

Answer 1

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 `ax^2+bx+c`.

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

`64x^2-1=0`
`64x^2=1`
`x^2=1/64`
`x=sqrt(1/64)`
`x=(sqrt(1))/sqrt(64)`
`x=+-1/8`

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

You have your roots already: `color(green)(x=-1/8)` and `color(red)(x=1/8)`.

Let's just equal each to zero:

`color(green)(8x+1=0)`
`color(red)(8x-1=0)`

Now, you know that

`64x^2-1=(8x+1)(8x-1)`

Answer 2

`64x^2 = (8^2)x^2=(8x)^2`, so it is a perfect square.

`1 = 1^2`, so it is also a perfect square.

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

`a^2-b^2 = (a+b)(a-b)`

So,

`64x^2-1 = (8x)^2 - (1)^2 = (8x+1)(8x-1)`

Now solving `color(white)"sssssssss"` `64x^2-1 = 0`

is the same as solving `(8x+1)(8x-1) = 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 `(8x+1)(8x-1) = 0`, we must make either:

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

We can make `8x+1 = 0`, by `8x = -1` so `x = -1/8`

And we can make `8x-1 = 0` by `8x = 1` so `x = 1/8`

The solutions are: `-1/8` and `1/8`

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

`64x^2-1 = 0`

`(8x+1)(8x-1) = 0`

`8x+1=0` `color(white)"sss"` or `color(white)"sss"` `8x-1=0`

`8x=-1` `color(white)"sss"` or`color(white)"sss"` `8x= 1`

`x=-1/8` `color(white)"sss"` or`color(white)"sss"` `x= 1/8`

The solutions are: `-1/8` and `1/8`