How do you solve using the completing the square method `x^2 - 5x = 9`?

Completing the square means seeking a constant term `n` to add to `x^2-5x`, so that `x^2-5x+n` is a perfect square.

First, let's look at what happens when we FOIL a perfect square binomial of the form `(x+a)^2`:

`(x+a)(x+a)=x^2+2ax+a^2`

A perfect square will always have a distributed form like this.

What we notice is that if we take the coefficient of the `x` term (`2a`), cut it in half, and then square it, we get `a^2`, the constant term. Thus, if given `x^2+2ax=b`, we would complete the square by adding `a^2` (that is, the square of half of `2a`) to both sides, giving

`x^2+2ax+a^2 = b + a^2`

so that the trinomial on the left will be guaranteed to be a perfect square—the square of `(x+a)`.

For this particular problem, we are given `x^2-5x=9`. So, `-5` is like our "`2a`". And if

`-5=2a`,

then

`a=-5/2`,

and

`a^2=25/4`.

Thus, `x^2-5x+25/4` is the completed square we seek, meaning we need to add `25/4` to both sides:

`x^2-5x+25/4=9+25/4`

Okay, so if this `x^2-5x+25/4` is a perfect square, what is its factored form? (Or, what is its square root?)

That's easy—remember that the factored form of `x^2+color(red)(2a)x+a^2` is `(x+color(blue)a)^2`. The `color(blue)a` in the factor is half of the `color(red)(2a)` in the trinomial. So the factored form of `x^2-5x+25/4` will be `(x-5/2)^2`, because `-5/2` is half of `-5`.

We simplify both sides now to get

`(x-5/2)^2=61/4`

Now our LHS is a perfect square, so we can solve for `x` by taking the square root of both sides:

`x-5/2=+-sqrt61 /2`

and then adding `5/2` to both sides:

`x=5/2+-sqrt61 / 2" "=" "(5+-sqrt61)/2`.

Note:

If the coefficient on the `x^2` term is something other than `1`, you'll want to either factor it out or divide everything by it first, so that this method will work.