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

2 Answers
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 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)

May 20, 2015

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" orcolor(white)"sss" 8x= 1

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

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