Question

How do you solve `2x^2 -2x- 2= 0` using completing the square?

Answer 1

Quick answer:
`2x^2-2x-2=0`
`2x^2-2x = 2`
`x^2 - x = 1`
`x^2 - x + (1/2)^2= (1/2)^2+ 1`
`(x-1/2)^2 = 5/4`
`x-1/2 = +-sqrt5/2`
`x=(1+-sqrt5)/2`

But what did we do and why?

In order to have a habit for how we do this, let's begin by moving the constant to the other side of the equation. We'll add `2` to both sides to get:

`2x^2-2x = 2`

When completed, the square will be on the left, and it will have the form:

`x^2+-2ax+a^2`,

so the next thing to do is get a `1` (which we won't write) in front of the `x^2`. (In fancy terms, we're going to make the coefficient of `x^2` equal to `1`.)

Multiply both sides of the equation by `1/2` (Remember to distribute the multiplication.)

`1/2(2x^2-2x) = 1/2(2)` now simplify:

`x^2 - x = 1`

Now that we have just `x^2`, we can see that the coefficient of `x` is negative, that tells us that the completed square will look like:

`x^2-2ax+a^2` which we will be able to factor: `(x-a)^2`

We have: `x^2 - x = 1`,
Which we can think of as: `x^2 - 1x = 1`,

so we must have:

`2a = 1`.

And that makes `a=1/2`. Square that and add the result to both sides:
`(1/2)^2 = 1^2/2^2 = 1/4`

`x^2 - x +1/4= 1/4+ 1`,

Now factor on the left (we already know how to factor that! See above.) And add on the right.

`(x-1/2)^2 = 1/4+4/4 = 5/4`

`(x-1/2)^2 = 5/4`

Now use the fact that `n^2 = g` if and only of `n = sqrtg or -sqrtg`
(The square of a number equals a given number if and only if the number is either the positive or negative square root of the given.)

`x-1/2 = +-sqrt(5/4) = +-sqrt5/sqrt4 = +-sqrt5/2`

`x-1/2 = +-sqrt5/2` Add `1/2` to both sides:

`x=1/2 +- sqrt5/2` which we often prefer to write as a single fraction:

`x=(1+-sqrt5)/2`

Note: never let yourself think this is some kind pf weird positive and negative number. It is just a convenient way of writing the two solutions:
`x=(1+sqrt5)/2` and `x=(1-sqrt5)/2`.