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

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

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Answer 1 · 2015-05-20T17:14:01

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

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

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

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

You have your roots already: $x = - \frac{1}{8}$ $color(green)(x=-1/8)$ and $x = \frac{1}{8}$ $color(red)(x=1/8)$ .

Let's just equal each to zero:

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

Now, you know that

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

Answer 2 · 2015-05-20T18:07:44

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

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

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

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

So,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$x = - \frac{1}{8}$ $x=-1/8$ $sss$ $color(white)"sss"$ or $sss$ $color(white)"sss"$ $x = \frac{1}{8}$ $x= 1/8$

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

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

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