Hi Ryan.

You can find several examples of completing the square by using the Quick Search and typing in completing the square.

I'm going to solve a similar problem: x² - 6x - 39 = 0.

Start by moving the scalar to the other side by negating it:
x² - 6x = 39

Then take half the co-efficient of the x term and square it: -6/2 = -3; (-3)² = 9. Add this to both sides:
x² - 6x + 9 = 48

Then you have "completed the square" so the expression on the left is a perfect square. Its root is x - 3, because 3 is one-half the x term's co-efficient (it is minus, because we have - 6x). So we re-write it as
(x - 3)² = 48

Now take the square root of both sides, remembering that a positive number on the right has both positive and negative roots. When you take the square root of a perfect square on the left side, the square root cancels with the squared:
x - 3 = ±√48

Then finish by isolating x by itself (move the 3 to the other side by negating) and simplifying the √48:
x = 3 ± 4√3

Notice that this is two answers for x: Solution set = { 3 - 4√3, 3 + 4√3 }.

On an exam, if I have time, I will check my answer as follows:
Let x = 3 - 4√3
Then x² - 6x - 39
= (3 - 4√3)² - 6(3 - 4√3) - 39
= (9 -24√3 + 16(3) ) - 18 + 24√3 - 39
= 9 - 24√3 + 48 - 18 + 24√3 - 39
= 0, as we expected.
Then I'd check the other answer for x the same way.

Always try our search before posting a question - it saves you time.

Cheers,
Stephen La Rocque.