Going to have a look at a very simple process. It's called completing the square. In order to get to it and to show its potential use, I want to start with a simple equation. X squared equals 9. In order to find out what taxes we would take the square root of both sides, so the square root of X squared is just X and the square root of 9 is plus three or minus three. Be'cause minus three all squared is also 9. So that was relatively straightforward. Both of these two numbers were square numbers, complete squares. What if we got? X squared is equal to 5. Do the same again, so we take the square root of each side X equals. Now 5 is not a square number, it does not have an exact square root, but nevertheless we can write it as root 5 that is exact or minus Route 5. So far so good. The same process is working each time. Let's have a look at something now, like X minus 7. All squared equals 3. How can we solve this? Well again, this side of the equation. We've got something called a complete square, so we can take the square root of each side. X minus Seven is equal to the square root of 3 or minus the square root of 3, and so we can now add 7 to each of these. So X is equal to 7 Plus Route 3 or 7. Minus Route 3. I just have a look at another One X plus three, all squared is equal to 5. Again, this is a complete square, so again we can take the square root X +3 is equal to Route 5 or minus Route 5. Now to get X on its own, we need to take three away from each side, so we have X equals minus 3 + 5 or minus 3 minus Route 5. What if we got X squared plus six X equals 4? Problem is this X squared plus 6X is not a complete square, so we can't just take the square root. So in terms of handling something like this, we've got to have a way of getting a complete square. So the process that we're going to be looking at it's called completing the square. The sorts of expressions that we have before will like this X plus a all squared. Or X minus a all squared, so that's what a complete square looks like. One of these two. So let's multiply this out and see what we get. So this is X plus a times by X plus A. And we do X times by X. That gives us X squared. And we do a Times by X, which gives us a X. And then with X times by a, which again gives us a X and then at the end a Times by a, which gives us a squared. And so we've X squared plus 2X plus A squared. I can do the same with this One X minus a Times by X minus A and it's going to give very similar results X times by X will give me X squared X times by minus A minus 8X minus 8 times by X minus 8X minus 8 times by minus A plus a squared. So tidying up the two middle bits. Minus X minus X, minus 2X and then plus A squared. So this is what complete squares look like. They look like one of these two. Well. Can I make this look like a complete square in some way shape or form? If I compare this with this, what is it that I see? Well, perhaps one of things I might like to have is this written as just a function X squared plus 6X minus four? And let's not worry too much about solving an equation. What we want to concentrate on is this process of completing the square, so I'm going to take this quadratic function X squared plus 6X minus four, and I'm going to compare it with the complete square. X squared plus 2X plus A squared. Now The X squared so the same. 6X2A X I've got to have these two terms the same. They've got to match. They've got to be exactly the same, and that means that the six has to be equal to 2A. And of course, that tells us that the A is equal to 3. So if I make a equal to three, then I've got plus A squared on the end +9. So. I can look at this first bit and I can make it equal to that. So let's write this down X squared plus 6X minus 4 equals. X plus three all squared. Remember this is X plus three all squared. Now I'm replacing the A by three. Now what more of I got? Well, I've added on a squared, so I've added on 9. So I've got some how to get rid of that. Well, let's just take it away, minus 3 squared. And then I can keep this minus four at the end as it was. So now I've X plus three all squared minus 9 - 4 gives me X plus three all squared minus 30. So I have completed the square. I made this bit. Part of a complete square. And I've done it by comparing the coefficient of X. With one of the two standard forms and I saw that what I had to do was take half the coefficient of X. Let's have a look then at another example, X squared minus 8X plus Seven and I want to write this so it's got a complete square in it. Well, one of the standard forms for the complete square that we had was X squared minus 2X plus A squared. And I want to make these two terms. The same. So again, we can see that the A. Has got to be 4 because minus 8 is minus 2 times by 4. So we've got X squared minus 8X plus Seven is equal to. X minus four all squared. So I've ensured I've got the X squared. I've ensured that I've got the minus 8X, but I've also added on a squared, so I've got a squared too much, so I must take away 4 squared and then I've got the 7:00 that I need to add on to keep the equal sign. And so this is now X minus four all squared minus 16 + 7 X minus four all squared minus 9. Let's take one more example. X squared Plus 5X plus three and let's see if we can follow this one through without having to write down the comparison. In other words, by doing it by inspection. What do we need? We need a complete square, so we need X and we look at this number here. The coefficient of the X turn on this left hand side. We want half that coefficient, so we want five over 2 and we've got a plus sign, so it's got to be X +5 over 2. If we were to multiply out this bracket, we would be adding on an additional A squared where five over 2 is the a. So we've got to take that away. Takeaway 5 over 2 squared and then at the end we've got plus 3. So this gives us X +5 over 2 or squared minus 25 over 4 + 3. And of course, we'd like to combine these numbers at this end X +5 over 2 all squared minus 25 over 4 plus. Now we need to convert these to quarters as well, so three is 12 quarters. Now we can combine these, since they're both in terms of quarters X +5 over 2, all squared minus 13 over 4. And we'd leave you like that. This process now seems to be working quite well. But of course, we haven't dealt with every kind of quadratic expression we could have, because so far we've only had a unit coefficient here in front of the X squared. We haven't had another number like two or three or whatever, so let's have a look at what we would do in that case. So we have three X squared. Minus 9X. Plus 50, what do we need to do to begin with? Well, we know how to do this if we've got a unit coefficient with the X squared, so let's make it a unit coefficient by taking out the three as a common factor. So that's three brackets X squared, minus 3X. Now 50. What are we going to do with this? Well? We divide it by three in order that when we do the multiplication 3 * 50 over three will just give us back the 50. Now we look at this thing here in the bracket because this is now exactly the same sort of expression that we've had before. Equals 3. Let's have a big Curly bracket. I'm going to make this going to complete the square around this, so this is going to be X minus. Look at the coefficient of X and take half of it 3 over 2 all squared. By doing that, we've added on. An additional A squared, so we need to take that off 3 over 2 squared and then finally plus 50 over 3 and close the big bracket. 3. X minus three over 2 all squared minus nine over 4 + 50 over 3 and closed the big bracket. Now all we need to do now is put these together. And to do that we need a common denominator and the common denominator. Four and three is going to be 12. 3 the Big Curly Bracket X minus three over 2 all squared minus over 12. We need to change the nine over 4 into 12. 3/4 gave us 12 so 3 nines give us 27. Plus we need to change the 50 over 3 into twelfths. 4 * 3 is 12, so 4 * 50 is 200. And now I've got a little bit of arithmetic to do. Let's just write back bracket down again. Equals 3. The Curly bracket. X minus three over 2 all squared. Minus 27 over 12. Plus 200 over 12 and that's the calculation that we need to do here to simplify 3 Curly bracket X minus three over 2 or squared. Los all over 12 and we're going to take the 27 away from the 200 is going to give us 173 and then we can close the bracket off. So despite the fact that the numbers were quite fearsome there, we've still ended up with a complete square, and we've automated the process so that what we're doing is looking at the coefficient of X. First of all, we check that what we've got the coefficient of X squared is one. If it's not, we take out the coefficient of X squared as a factor. Next we check the coefficient of X and we take a half of it, and that's the number that's going to go here inside the bracket. Then we must remember that we've got take off the square of that number is effectively we've added it back on, and then the rest is just arithmetic. So now we've developed this technique of completing the square. Let's use it to solve our original problem. If you remember we had X squared plus 6X is equal to four and we chose to write that as X squared plus 6X minus 4 equals 0. So first we need to check has it got a unit coefficient. And it has, so we don't need to take out a common factor. Now we look at the coefficient of X and it's 6 and it's half that coefficient that we want. So we need 3. So this is going to be X. Plus three all squared. In doing that, we have added on 3 squared, so we need to take off that 3 squared. And now we need to include the minus 4. So that we can maintain the quality of this xpression with this one and it's equal to 0. So X plus three all squared. Minus 3 squared. That's minus nine and minus 4 equals 0, so we can combine these X plus three all squared minus 13 equals not. At the 13 to each side X plus three, all squared equals 13, and now we're in a position to take the square root of both sides. Because here on the left hand side we have a complete square. And so this is X +3 equals. Now 13 isn't a complete square. It's not a square number, so we have to write it as square root of 13. Or remembering when we take a square root of a number it's plus or minus Route 30. Now we take the three away from each side and we end up with our two roots. So we take the three away we have minus 3 + 13. Or minus 3 minus Route 30 and so that process of completing the square can be used to help us solve a quadratic equation. But that's not the real issue here. You can see another video on solving quadratic equations. The point is to master this technique of completing the square.