 One of the standard methods of solving quadratic equations is known as completing the square, and we'll take a little bit of an overview of that. If I have a quadratic equation like x squared plus 10x equal to 39, I have a number of different ways that I can use to try and solve this equation. So one possibility is I might try to factor. And in fact, this is a fairly common way of solving it. There's just one problem with it, which is that factoring is painful, tedious, and in many cases, it's actually not possible. And if I'm looking at a real problem, in general, I will not be able to factor the equation that arises from a realistic problem. So what I can do is I can always fall back on the quadratic formula, and this is fast, efficient, it always works. But as a third method, I can do what's called completing the square. Now this is somewhat tedious, not quite as bad as factoring, but it's still a tedious procedure. However, the reason that it's important is that completing the square is actually useful for other things. So for example, let's consider this equation x squared plus 10x equals 39. And one way to look at this is to use geometry to represent each side of the equation. And how we'll do that is we'll let these terms correspond to the area of some figure. Now it's even more helpful if we break apart our expressions into individual terms. So for example, this left hand side, I have an x squared, I have a 10x. If I take a look at these two terms individually, I can try and draw a geometric picture of what they look like. So let's think about that. That x squared term, if I want to represent these terms by the area of some rectangle, well I know the area of a rectangle is the product of length and width. So I need to find two quantities, length and width, whose product is going to be x squared. And after thinking about that, the first thing that leaps to mind is the quantity 1 over square root of 253x cubed and square root of 253 times x to the fit. Maybe that's not the first thing that you want to think about. Maybe a little bit better is to use the quantities x and x. So if I multiply length x by width x, I get area x squared. Now because the length and width are equal, the rectangle that I produce here is actually going to be a real square. Likewise, my other term, 10x, I can view this as a rectangle, that's 10 by x. And here's the important thing, one side of that rectangle has with x. So what that means is I can take my square and I can attach this 10 by x rectangle to the end of the square, and I produce a larger figure that looks something like that. And now if I want to think about this problem as something to solve, I have x squared plus 10x equals 39. And so that's this figure here has an area of 39. And I can solve this equation if I can find the length and width of the figure. So there's just one problem with that. I need to find two numbers that multiply to 39. So maybe I have 39 by 1. Well I could actually have other numbers as well. Maybe it's 1 by 39. So maybe this is with 1 length 39. Or possibly maybe it's a 3 by 13 or a 13 by 3. Or maybe a one-fifth by 195 or many, many other possibilities. And the problem is there's too many possibilities for us to figure out which one is the one that we want. Well the problem is that given the area of a rectangle, given that I know the area of this rectangle is 39, I'm not going to be able to find the lengths of the two sides of the rectangle. Unless I happen to have a square. The problem is what I have is not a square. It's x by x and then some. So this thing is definitely not a square. Well what I want to do is I want to make it a square. And this is why this method is actually called completing the square. Now to do that what we're going to do is we're going to take this extension. We're going to cut it in half. And then well what's the key property of a square? The length and width are equal. And so here I originally had an x plus 10 width by x height. I want to make this at least have equal width and height. So what I'm going to do is after I cut this piece in half I'm going to slide this piece over and attach it to the bottom there. Now part of what makes this work is that this piece here is a 5 by x rectangle. So I know that a 5 by x rectangle can fit on the bottom of this square. And what I have is this L shaped region that's called a nomen. And it's the same as the original figure. It's just been moved around a little bit. So the area of this region is still equal to our original value of 39. So maybe I'll just make a note of that. The area of this is equal to 39. And while I'm missing a piece this would be a square x plus 5 by x plus 5. If I had this missing piece here. Well I could go ahead and put that there. The problem is I don't know what the area of that piece is. If only I knew how big this square was. Oh wait a minute. If this length is 5 so is this length. And if this length is 5 so is this length. So this piece here is a square that's 5 by 5. And I know how to calculate the area of that. The missing piece has area 25. And that tells me the whole thing. The complete square has area 39 plus 25. The area is going to be 64. And again once I know the area of the square. I know the length of each side of the square. The square root literally the length of the side of a square with a given area. That square root is going to be 8. So that tells me these sides have to be length 8. And that also tells me that x plus 5 is 8. So I know that x is equal to 3. Well when we go from geometry to algebra. We actually have a slightly more expanded repertoire. So let's see what happens. If we go through the same steps to solve our x squared plus 10x equals 39. I had my square x squared. I had my 10x. And then I split it. I shifted pieces around. That doesn't change anything. The area is still equal to 39. And this time I added that little green square here area 25. And what I got over on the right hand side 39 plus 5 squared 64. Over on the left hand side I have the square of x plus 5. Literally the square of x plus 5. And so that tells me x plus 5 is plus or minus the square root of 84. Because we're looking at algebra. We're no longer restricted to having positive lengths. So I have to indicate that by pointing out that we could have a plus square root of 64. Or a minus square root of 64. And that tells me that x is plus or minus square root of 64 minus 5. Which is to say it's either 8 minus 5 or it's negative 8 minus 5. And my two solutions 3 and negative 13. Now I know some of you are saying oh my god that's a really complicated picture. I don't want to have to draw this horribly complicated picture here. Can I do this without drawing pictures? And the answer is yes. But you have to be a little bit careful. So let's go ahead and take a look at how this might look without drawing the pictures. So here's my equation x squared plus 14 x equals 50. And again if we go through our process of what we did. We split the x coefficient in half. We squared it. We added the squares. We did a little bit of arithmetic over on the left hand side. We have perfect square x squared plus 7 squared. Over on the right hand side we have something we can do arithmetic on. I take the square root of both sides. Plus or minus because I'm living in algebra land. And then I'm going to subtract that 7. And I end up with my two solutions.