Worked example: Solving equations by completing the square | High School Math | Khan Academy

So let's see if we can solve this quadratic equation right over here: (x^2 - 2x - 8 = 0).

And actually, they're cutting down some trees outside, so my apologies if you hear some chopping of trees. Well, I'll try to ignore it myself.

All right, so back, back, back to the problem at hand. There's actually several ways that you could attack this problem. You could just try to factor the left-hand side and go that way, but the way we're going to tackle it is by completing the square.

And what does that mean? Well, that means that I want to write the left-hand side of this equation in the form ((x + A)^2 + B). As we'll see, if we can write the left-hand side in this form, then we can actually solve it in a pretty straightforward way.

So let's see if we could do that. Well, let's just remind ourselves how we need to rearrange the left-hand side in order to get it to this form. If I were to expand out ((x + a)^2), let me do that in a different color.

So, if I were to expand out ((x + a)^2), that is (x^2 + 2ax + a^2) — make sure that plus sign, you can see it: plus (2ax) plus (a^2), and of course, you still have that plus (B) there.

So let's see if we can write this in that form. What I'm going to do — and this is what you typically do when you try to complete the square — I'll write (x^2 - 2x), now I'm going to have a little bit of a gap.

And I'm going to have (-8), and I have another little bit of a gap, and I'm going to say equals (0). So I just rewrote this equation, but I gave myself some space so I can add or subtract some things that might make it a little bit easier to get into this form.

So if we just match our terms (x^2), (x^2), (2ax) equals (-2x). So if this is (2ax), that means that (2a = -2). So (a = -1).

Another way to think about it, your (a) is going to be half of your first-degree coefficient, or the coefficient on the (x) term. So the coefficient on the (x) term is (-2); half of that is (-1).

And then we want to have (a^2), so if (a) is (-1), (a^2) would be (+1). So let's throw a plus one there. But like we've said before, we can't just willy-nilly add something on one side of the equation without adding it to the other or without subtracting it again on that same side; otherwise, you're fundamentally changing the truth of the equation.

So if I add one on the left side, I either have to add one on the right side to make the equation still hold true, or I could add one and subtract one from the left-hand side. So I'm not really changing the value of the left-hand side; all I've done is added one and subtracted one from the left-hand side.

Now, why did I do this again? Well, now I've been able — I haven't changed its value — I just added and subtracted the same thing. But this part of the left-hand side now matches this pattern right over here: (x^2 + 2ax) where (a) is (-1).

So it's (x^2 - 2x + 1^2), and then this part right over here is the plus (B). So we already know that (B = -8 + 1 = -7), and so that's going to be our (B) right over there.

We can rewrite this as what I squared off in green: that's going to be ((x - 1)^2 - 9 = 0). Then I can add (9) to both sides.

So I just have this squared expression on the left-hand side. So let's do that; let me add (9) to both sides. And what I'm going to be left with — so let me just, on the left-hand side, those cancel out; that's why I added the (9) — I'm just going to be left with ((x - 1)^2 = 9).

So if ((x - 1)^2 = 9), that means that that something is either going to be the positive or the negative square root of (9).

So it's either going to be (x - 1 = 3) or (x - 1 = -3). And you could see that here: if (x - 1 = 3), (3^2 = 9). If (x - 1 = -3), then ((-3)^2 = 9).

So here we can just add (1) to both sides of these equations. Add (1) to both sides of this equation, and you get (x = 4) or (x = -2).

So, (x) could be equal to (4) or (x) could be equal to (-2), and we're done.

Now, some of you might be saying, "Well, why did we go through the trouble of completing the square? I might have been able to just factor this and then solve it that way."

And you could have actually; for this particular problem, completing the square is very powerful because you can actually always apply this. In the future, you will learn the quadratic formula, and the quadratic formula actually comes directly out of completing the square.

In fact, when you're applying the quadratic formula, you're essentially applying the result of completing the square.

So hopefully you found that fun!