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