Question

How do you factor and solve `2x^2 - 3= 125`?

Answer 1

`x=+-8`

Explanation:

`2x^2-3=125`

Subtract 125 on both sides

`2x^2-128`=0

Divide both sides by 2

`x^2-64=0`

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

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

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

`x=+-8`

Answer 2

`2x^2-3=125` can be factored to:

`2(x-8)(x+8)=0`, and has the solution:

`color(red)(absx=8)`

Explanation:

Move all the terms to one side of the equation

`2x^2-3=125`

`2x^2-3-color(red)125=cancel125-cancelcolor(red)125`

`2x^2-128=0`

Now take out a factor of 2

`(color(red)2*x^2)-(color(red)2*64)=0`

`color(red)2(x^2-64)=0`

We now have a term in the parentheses that looks like

`(a^2-b^2)`

This is called a difference of squares

We can factor a difference of squares like this:

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

Let's apply this to our expression

`2(x^2-color(red)64)=0`

`2(x^2-color(red)(8^2))=0`

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

This is the fully factored form.

By examining this equation, we can see that the solutions — the values of `x` that make the equation true — are

`x=8`

and

`x=-8`

or simply

`absx=8`