 Hello everyone, today I just want us to go over a few problems dealing with completing the square. So in class we talked about a lot of the different properties and patterns that arise with completing the square. And I just want to go through one example in which we solve an equation by completing the square. So the equation we're going to look at is x squared plus 8x minus 3. And the first thing that you need to do when solving by completing the square is move the constant to the other side of the equation. So you want all of your x's on one side and then any constants on the other side. The reason we can do that is now we have an equation set up so that we can add something to each side and make this left side of the equation a perfect square. And if you remember from what we did in class, if we look at this b value, the number in front of the x, and divide it by 2, we would get 4, and then square it to get 16. That will make a perfect square on the left side of the equation. So now this part in brackets here is a perfect square and what that means is we can write it as x plus or minus something squared. In this case we can again look at the b value, the 8, and divide it by 2, which is a positive 4, and that completes our square. x plus 4 squared is the same as x squared plus 8x plus 16. So we now have a perfect square and so our equation is x plus 4 squared equals 19. Now to solve this you can just use square roots. So our square root property says we can take the square root of each side of the equation. On the left we have x plus 4, on the right we have a positive or negative square root of 19. And then the final step to solve for x will be to subtract 4, and what that gives us is x equals negative 4 plus or minus the square root of 19. Typically, since these are both real numbers, I would just evaluate them each separately. So 4 plus, excuse me, negative 4 plus the square root of 19 is 0.359. So that's one of our solutions and negative 4 minus the square root of 19 is negative 8.359. And so our two solutions to that problem are x equals 0.359 and x equals negative 0.8359. Take a minute and pause the video to try the next few problems, numbers 10 through 12. If you do get stuck on any of the problems feel free to post a question to the discussion board or email me. Now I want to go through one example of completing the square when the leading coefficient is not equal to 1. We will use the same process that we used before, but there will just be one added step. So just like before, what we're going to do first is move any constants to the other side of the equation. In this case we'll add 2 to each side in order to get the equation 3x squared plus 5x equals 2. Next, in order to make a perfect square you need to make sure the leading coefficient is 1. In order to make it 1 you can divide everything by the leading coefficient. Here I'm dividing everything by 3 in order to give me this new equation. Now this is a little bit messier because of all the fractions, but we can complete the square in the same way as before. On the left we have the problem set up so that we can just add something to each side of the equation to make a perfect square. The way we figure out what to add is look at that b value, in this case the 5 thirds. So we're going to take 5 thirds divided by 2 and then square it. 5 thirds divided by 2 is 5 over 6 and so when we square 5 over 6 we get 25 over 36. On each side of the equation we will just add 25 over 36. Alright I'm going to get a new slide here and I'll rewrite what we had. Now on the left we now have a perfect square. So again it can be written as x plus or minus something squared. To determine what to add or subtract we again look at this b value divided by 2 and that gives us 5 over 6. So x plus 5 over 6 squared is equal to x squared plus 5 thirds x plus 25 over 36. Now on the left you just can add those together. If you want you can just get common denominators or you can type it in your calculator in order to find the fraction. But leaving it as a fraction will help us get an exact solution. So this is our new equation we're going to solve. And once the left side is written as a perfect square it now can be solved by taking the square root of each side. So on the left we're left with x plus 5 over 6. On the right we can square root the numerator and denominator separately so we have a positive or negative 7 over 6. The final step to solve for x will be to subtract 5 over 6 giving us a solution of negative 5 over 6 plus or minus 7, 6. If you evaluate that you get 2, 6 which is equal to 1 third or negative 12 over 6 which is equal to negative 2.